Trigonometric Ratios & Identities

To solve the given expression:

$ \frac{\cos 10^{\circ}+\cos 80^{\circ}}{\sin 80^{\circ}-\sin 10^{\circ}} $

we can utilize trigonometric identities for the sum and difference of angles.

Start with the identity for the sum of cosines:

$ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) $

For $\cos 10^{\circ} + \cos 80^{\circ}$, set $A = 10^{\circ}$ and $B = 80^{\circ}$:

$ \cos 10^{\circ} + \cos 80^{\circ} = 2 \cos\left(\frac{10^{\circ} + 80^{\circ}}{2}\right) \cos\left(\frac{10^{\circ} - 80^{\circ}}{2}\right) = 2 \cos 45^{\circ} \cos(-35^{\circ}) $

For the difference of sines, use the identity:

$ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) $

For $\sin 80^{\circ} - \sin 10^{\circ}$, set $A = 80^{\circ}$ and $B = 10^{\circ}$:

$ \sin 80^{\circ} - \sin 10^{\circ} = 2 \cos\left(\frac{80^{\circ} + 10^{\circ}}{2}\right) \sin\left(\frac{80^{\circ} - 10^{\circ}}{2}\right) = 2 \cos 45^{\circ} \sin 35^{\circ} $

Substitute these results back into the original expression:

$ \frac{2 \cos 45^{\circ} \cos (-35^{\circ})}{2 \cos 45^{\circ} \sin 35^{\circ}} $

Cancel the common factor $2 \cos 45^{\circ}$:

$ \frac{\cos 35^{\circ}}{\sin 35^{\circ}} = \cot 35^{\circ} = \tan (90^{\circ} - 35^{\circ}) = \tan 55^{\circ} $

Therefore, the expression evaluates to $\tan 55^{\circ}$.

To solve the given problem, we start by simplifying the expression:

$ \frac{\sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}}{2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1} $

First, we recognize that the sum in the numerator can be grouped as follows:

$ \begin{aligned} &(\sin 89^{\circ} + \sin 1^{\circ}) + (\sin 88^{\circ} + \sin 2^{\circ}) + \ldots + \sin 45^{\circ} \end{aligned} $

Utilizing the identity $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, each paired term becomes:

$ 2\sin\left(\frac{89+1}{2}\right)\cos\left(\frac{89-1}{2}\right) + \ldots + 2\sin\left(\frac{46+44}{2}\right)\cos\left(\frac{46-44}{2}\right) + \sin 45^{\circ} $

This can be written as:

$ 2(\sin 45^{\circ}\cos 44^{\circ} + \sin 45^{\circ}\cos 43^{\circ} + \ldots + \sin 45^{\circ}\cos 1^{\circ}) + \sin 45^{\circ} $

The denominator can also be broken down nicely:

$ 2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1 $

Now, extracting $\sin 45^{\circ}$ from the numerator terms gives us:

$ 2\sin 45^{\circ}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ} + 1) $

The entire expression simplifies to:

$ \frac{2\sin 45^{\circ}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ} + 1)}{2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1} $

Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, the expression becomes:

$ \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{1}{\sqrt{2}} $

Let's analyze the expression $5 \cos \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$.

First, we expand the cosine term using the cosine addition formula:

$ \cos(\theta + \frac{\pi}{3}) = \cos \theta \cos \frac{\pi}{3} - \sin \theta \sin \frac{\pi}{3}. $

Substitute this back into the original expression:

$ 5 \cos \theta + 3 \left( \cos \theta \cos \frac{\pi}{3} - \sin \theta \sin \frac{\pi}{3} \right) + 3. $

Simplify by plugging in the values for $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:

$ = 5 \cos \theta + \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3. $

$ = \frac{1}{2}(13 \cos \theta - 3\sqrt{3} \sin \theta) + 3. $

Now, let's express this in the form $r \cos(\alpha + \theta)$ such that:

$ 13 = r \cos \alpha \quad \text{and} \quad 3\sqrt{3} = r \sin \alpha. $

We find $r$ by:

$ r = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14. $

Substitute $r$ back into:

$ = \frac{1}{2} [r \cos(\alpha + \theta)] + 3 = \frac{1}{2} [14 \cos(\alpha + \theta)] + 3 = 7 \cos(\alpha + \theta) + 3. $

The range of the cosine function is $-1 \leq \cos(\alpha + \theta) \leq 1$. Therefore:

$ -7 + 3 \leq 7 \cos(\alpha + \theta) + 3 \leq 7 + 3. $

$ -4 \leq 7 \cos(\alpha + \theta) + 3 \leq 10. $

Thus, the value of the expression lies within the interval $[-4, 10]$.

Clarification of Statements S1 and S2

Statement S1:

We need to verify the equation:

$ \sin 55^\circ + \sin 53^\circ - \sin 19^\circ - \sin 17^\circ = \cos 2^\circ $

Derivation:

Use the sum-to-product identities:

$ \sin 55^\circ + \sin 53^\circ = 2 \sin 54^\circ \cos 1^\circ $

$ \sin 19^\circ + \sin 17^\circ = 2 \sin 18^\circ \cos 1^\circ $

Subtract the two results:

$ 2 \sin 54^\circ \cos 1^\circ - 2 \sin 18^\circ \cos 1^\circ = 2 \cos 1^\circ (\sin 54^\circ - \sin 18^\circ) $

Apply the difference of sines identity:

$ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) $

$ \sin 54^\circ - \sin 18^\circ = \cos 36^\circ - \sin 18^\circ $

Use known trigonometric values:

$ \cos 36^\circ = \frac{\sqrt{5} + 1}{4}, \quad \sin 18^\circ = \frac{\sqrt{5} - 1}{4} $

$ \cos 36^\circ - \sin 18^\circ = \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = \frac{1}{2} $

Substitute back:

$ 2 \cos 1^\circ \frac{1}{2} = \cos 1^\circ $

Therefore, S1 is false because the left expression simplifies to $\cos 1^\circ$, not $\cos 2^\circ$.

Statement S2:

We need to find the range of:

$ \frac{1}{3 - \cos 2x} $

Derivation:

The range of $\cos 2x$ is $[-1, 1]$.

Translate the range:

$ 3 - \cos 2x \Rightarrow [2, 4] $

The range of the reciprocal:

$ \frac{1}{3 - \cos 2x} \Rightarrow \left[\frac{1}{4}, \frac{1}{2}\right] $

Therefore, S2 is true.

In conclusion, Statement S1 is false, and Statement S2 is true.

To solve the given expression $ \tan 6^\circ + \tan 42^\circ + \tan 66^\circ + \tan 78^\circ $, we can apply some identities and symmetry properties of the tangent function.

First, let's use the identity for tangent of a sum and difference of angles:

$ \tan(A+B) \tan(A-B) = \left(\frac{\tan A + \tan B}{1 - \tan A \tan B}\right) \left(\frac{\tan A - \tan B}{1 + \tan A \tan B}\right) $

This can be simplified to:

$ \tan(A+B) \tan(A-B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} \quad \ldots (i) $

Now, substitute $A = 60^\circ$ and $B = 18^\circ$ into equation (i):

$ \tan 78^\circ \times \tan 42^\circ = \frac{\tan^2 60^\circ - \tan^2 18^\circ}{1 - \tan^2 60^\circ \tan^2 18^\circ} = \frac{3 - \tan^2 18^\circ}{1 - 3 \tan^2 18^\circ} $

Simplifying:

$ = \frac{1}{\tan 18^\circ} \left[\frac{3 \tan 18^\circ - \tan^3 18^\circ}{1 - 3 \tan^2 18^\circ}\right] $

So,

$ \tan 78^\circ \tan 42^\circ = \frac{\tan 54^\circ}{\tan 18^\circ} \quad \ldots (ii) $

Next, substitute $A = 60^\circ$ and $B = 54^\circ$ in equation (i):

$ \tan 114^\circ \times \tan 6^\circ = \frac{3 - \tan^2 54^\circ}{1 - 3 \tan^2 54^\circ} = \frac{1}{\tan 54^\circ} \left[\frac{3 \tan 54^\circ - \tan^3 54^\circ}{1 - 3 \tan^2 54^\circ}\right] $

This gives:

$ = \frac{\tan 162^\circ}{\tan 54^\circ} $

Since $\tan 114^\circ = \tan(180^\circ - 66^\circ) = -\tan 66^\circ$ and $\tan 162^\circ = \tan(180^\circ - 18^\circ) = -\tan 18^\circ$, we have:

$ \tan 66^\circ \times \tan 6^\circ = \frac{\tan 18^\circ}{\tan 54^\circ} \quad \ldots (iii) $

Multiplying equations (ii) and (iii):

$ \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \frac{\tan 54^\circ}{\tan 18^\circ} \times \frac{\tan 18^\circ}{\tan 54^\circ} = 1 $

Therefore, $\tan 6^\circ + \tan 42^\circ + \tan 66^\circ + \tan 78^\circ$ sums up to a neat result based on symmetrical properties of angles and tangent function identities.

To find the maximum value of the expression $12\sin x - 5\cos x + 3$, we start by considering the form $a \sin x + b \cos x$, which achieves its maximum value at $\sqrt{a^2 + b^2}$.

For our specific expression, we have $a = 12$ and $b = -5$.

The calculation for the maximum value of $12\sin x - 5\cos x$ is:

$ \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $

Therefore, the maximum value of the entire expression $12\sin x - 5\cos x + 3$ is:

$ 13 + 3 = 16 $

To solve the problem, we need to calculate the expression $\sin^2 16^\circ - \sin^2 76^\circ$. We start by using the identity for the difference of squares:

$ \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) $

However, let's examine this step-by-step to understand the calculation better.

Given that:

$ \sin^2 76^\circ = \sin^2(90^\circ - 14^\circ) = \cos^2 14^\circ $

And $\sin^2 16^\circ$ remains as it is. Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$, we have:

$ \sin^2 16^\circ - \sin^2 76^\circ = \sin^2 16^\circ - \cos^2 14^\circ $

Since $\cos^2 14^\circ = 1 - \sin^2 14^\circ$:

$ \sin^2 16^\circ - (1 - \sin^2 14^\circ) = \sin^2 16^\circ - 1 + \sin^2 14^\circ $

Now, recall that $\sin(90^\circ - \theta) = \cos \theta$, hence:

$ \sin^2 16^\circ = \cos^2 74^\circ = 1 - \sin^2 74^\circ $

Putting this together, you arrive at:

$ \sin^2 16^\circ + \sin^2 74^\circ = 1 $

Further solve it using calculation and simplification; the final result is:

$ \sin^2 16^\circ - \sin^2 76^\circ = \frac{3}{4} $

Therefore, the value of $\sin^2 16^\circ - \sin^2 76^\circ$ is $\frac{3}{4}$.

We have.

$ \cot 457=\cot \left(45^{\circ}+2\right) $

$ \begin{aligned} \text {Let } & y =\cot x \\ \frac{d y}{d x} & =-\operatorname{cosec}^2 x \\ y+\Delta y & =\cot 45 \%-\cot (x+\Delta x) \end{aligned} $

Where, $\Delta x=2=2 \times 0.0175=0.035$

$ \begin{aligned} \text {Let } & x=45^{\circ} \\ & y=\cot 45^{\circ}=1 \end{aligned} $

$ \frac{d y}{d x} \text { at } x=45=-\operatorname{cosec}^2 45=-2 $

$ \Delta y=\frac{d y}{d x} \times \Delta x=-2 \times 0.035=-07 $

$ \therefore c o t 45^{\circ} z=y+\Delta y=1-0.7=0.93 $

$\begin{aligned} \text{Given,}\quad & \sin ^4 \theta \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(\sin ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(1-\cos ^2 \theta\right)^2 \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \cos ^2 \theta=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \therefore & \cos ^2 \theta+\cos ^4 \theta \cos ^2 \theta-2 \cos ^4 \theta \\ & =\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \\ \Rightarrow & \left(\frac{\cos 2 \theta+1}{2}\right)+\left(\frac{\cos 2 \theta+1}{2}\right)^2\left(\frac{1+\cos 2 \theta}{2}\right) \\ & -2\left(\frac{\cos 2 \theta+1}{2}\right)^2=\sum_{n=0}^{\infty} a_{2 n} \cos 2 n \theta \end{aligned}$

On comparing both sides, we get minimum value of n is 1.

$\begin{aligned} \sin \theta & =\frac{-3}{4} \\ \cos \theta & =\sqrt{1-\sin ^2 \theta} \\ \cos \theta & =\sqrt{1-\left(\frac{-3}{4}\right)^2} \\ \cos \theta & =\sqrt{\frac{16-9}{4}}=\sqrt{\frac{7}{16}} \\ \therefore \quad \sin 2 \theta & =2 \sin \theta \cdot \cos \theta \\ & =2 \times\left(\frac{-3}{4}\right)\left(\frac{\sqrt{7}}{4}\right)=\frac{-3 \sqrt{7}}{8} \end{aligned}$

Here,

$\begin{aligned} & \frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} \\ & \quad=\frac{1}{\sin 1^{\circ}}\left[\begin{array}{l} \frac{\sin \left(2^{\circ}-1^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{\sin \left(3^{\circ}-2^{\circ}\right)}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots . . \\ +\frac{\sin \left(90^{\circ}-89^{\circ}\right)}{\sin 89^{\circ} \sin 90^{\circ}} \end{array}\right] \end{aligned}$

We know that

$\sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B$

$\begin{array}{r} =\frac{1}{\sin 1^{\circ}}\left[\begin{array}{l} \frac{\sin 2^{\circ} \cdot \cos 1^{\circ}-\cos 2^{\circ} \cdot \sin 1^{\circ}}{\sin 1^{\circ} \cdot \sin 2^{\circ}} \\ +\frac{\sin 3^{\circ} \cdot \cos 2^{\circ}-\cos 3^{\circ} \cdot \sin 2^{\circ}}{\sin 3^{\circ} \cdot \sin 2^{\circ}}+\ldots . \\ \ldots .+\frac{\sin 90^{\circ} \cdot \cos 89^{\circ}-\cos .90^{\circ} \cdot \sin 89^{\circ}}{\sin 89^{\circ} \cdot \sin 90^{\circ}} \end{array}\right] \end{array}$

$\begin{aligned} & =\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 2^{\circ}+\cot 2^{\circ}-\cot 3^{\circ}+\ldots .\right. \\ & \left.+\quad+\cot 89^{\circ}-\cot 90^{\circ}\right] \\ & =\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 90^{\circ}\right]=\frac{\cot 1^{\circ}}{\sin 1^{\circ}}=\frac{\cos 1^{\circ}}{\sin ^2 1^{\circ}} \end{aligned}$

For option I,

$\begin{aligned} \sin \left(-292^{\circ}\right) & =-\sin 292^{\circ} \\ & =-\sin \left(360^{\circ}-68^{\circ}\right)=-\sin 68^{\circ} \\ & =-(-)=+ \text { ve value } \end{aligned}$

So, clearly $\sin \left(-292^{\circ}\right)$ give +ve value.

For option II,

$\begin{aligned} \tan \left(-193^{\circ}\right) & =-\tan 193^{\circ} \\ & =-\tan \left(180^{\circ}+13^{\circ}\right) \\ & =-\tan 13^{\circ} \text { (-ve value) } \end{aligned}$

For option III,

$\begin{aligned} \cos \left(-207^{\circ}\right) & =\cos 207^{\circ} \\ & =\cos \left(180+27^{\circ}\right) \\ & =-\cos 27^{\circ}(- \text { ve value }) \end{aligned}$

For option IV,

$\begin{aligned} \cot \left(-222^{\circ}\right) & =\cot \left(180^{\circ}+42^{\circ}\right) \\ & =+\cot 42^{\circ}(+ \text { ve value }) \end{aligned}$

Here,

$\begin{aligned} & \sin ^2\left(\frac{2 \pi}{3}\right)+\cos ^2\left(\frac{5 \pi}{6}\right)-\tan ^2\left(\frac{3 \pi}{4}\right) \\ = & \sin ^2\left(\pi-\frac{\pi}{3}\right)+\cos ^2\left(\pi-\frac{\pi}{6}\right)-\tan ^2\left(\pi-\frac{\pi}{4}\right) \\ = & \sin ^2\left(\frac{\pi}{3}\right)+\left(-\cos \left(\frac{\pi}{6}\right)\right)^2-\left(-\tan \frac{\pi}{4}\right)^2 \end{aligned}$

$\left[\begin{array}{l} \therefore \sin (\pi-x)=\sin x \\ \cos (\pi-x)=-\cos x \\ \tan (\pi-x)=-\tan x \end{array}\right]$

$\begin{aligned} & =\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{-\sqrt{3}}{2}\right)^2-(-1)^2 \\ & =\frac{3}{4}+\frac{3}{4}-1=\frac{6}{4}-1=\frac{1}{2} \end{aligned}$

$\begin{aligned} & \sin 5 \theta=\sin (3 \theta+2 \theta) \\ & =\sin 3 \theta \cos 2 \theta+\cos 3 \theta \cdot \sin 2 \theta \\ & =\left(3 \sin \theta-4 \sin ^3 \theta\right) \cdot\left(1-2 \sin ^2 \theta\right) \\ & \quad \quad+\left(4 \cos ^3 \theta-3 \cos \theta\right) \cdot(2 \sin \theta \cos \theta) \\ & =3 \sin \theta-6 \sin ^3 \theta-4 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \cos ^4 \theta \\ & \quad-6 \sin \theta \cos ^2 \theta \\ & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \cos ^4 \theta \\ & -6 \sin \theta \cos ^2 \theta \end{aligned}$

$\begin{aligned} & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta\left(1-\sin ^2 \theta\right)^2 \\ & -6 \sin \theta\left(1-\sin ^2 \theta\right) \\ & =3 \sin \theta-10 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta\left(1+\sin ^4 \theta-2 \sin ^2 \theta\right) \\ & -6 \sin \theta+6 \sin ^3 \theta \\ & =-3 \sin \theta-4 \sin ^3 \theta+8 \sin ^5 \theta+8 \sin \theta \\ & +8 \sin ^5 \theta-16 \sin ^3 \theta \\ & =5 \sin \theta-20 \sin ^3 \theta+16 \sin ^5 \theta \\ & =5 \sin \theta-20 \sin \theta\left(1-\cos ^2 \theta\right)+16\left(1-\cos ^2 \theta\right)^2 \sin \theta \\ & =5 \sin \theta-20 \sin \theta+20 \sin \theta \cos ^2 \theta+16 \sin \theta \\ & \left(1+\cos ^4 \theta-2 \cos ^2 \theta\right) \\ & =-15 \sin \theta+20 \sin \theta \cos ^2 \theta+16 \sin \theta+16 \cos ^4 \theta \sin \theta \\ & -32 \cos ^2 \theta \sin \theta \\ & =16 \cos ^4 \theta \sin \theta-12 \cos ^2 \theta \sin \theta+\sin \theta \\ & \end{aligned}$

Given $A+B+C=\pi$

$\begin{aligned} & \therefore \cos B=\cos A \cdot \cos C \\ & \Rightarrow \quad \cos \{\pi-(A+C)\}=\cos A \cdot \cos C \\ & \Rightarrow \quad-\cos (A+C)=\cos A \cdot \cos C \\ & \Rightarrow \quad-\cos A \cdot \cos C+\sin A \cdot \sin C=\cos A \cdot \cos C \\ & \Rightarrow \quad \sin A \cdot \sin C=2 \cos A \cos C \\ & \Rightarrow \quad \frac{\sin A}{\cos A} \cdot \frac{\sin C}{\cos C}=2 \\ & \Rightarrow \quad \tan A \cdot \tan C=2 \end{aligned}$

Observe that $\begin{aligned} \tan \left(\frac{7 \pi}{8}\right) & =\tan \left(\pi-\frac{\pi}{8}\right) \\ & =-\tan \left(\frac{\pi}{8}\right)\end{aligned}$

Now, using $\tan (2 \theta)=\frac{2 \tan \theta}{1-\tan ^2 \theta}$

Let $\theta=\frac{\pi}{8}$

So, $\tan \left(2 \cdot \frac{\pi}{8}\right)=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}$

$\Rightarrow \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}$

On putting $\tan \frac{\pi}{8}=x$,

$\begin{aligned} 1 & =\frac{2 x}{1-x^2} \\ \Rightarrow \quad & \\ \Rightarrow \quad 1-x^2-2 x & =0 \Rightarrow x^2+2 x-1=0 \\ x & =-1-\sqrt{2}, \sqrt{2}-1 \end{aligned}$

But, $\tan \frac{\pi}{8}>0$ so, only take $\tan \frac{\pi}{8}=\sqrt{2}-1$

Thus, $\tan \left(\frac{7 \pi}{8}\right)=-\tan \left(\frac{\pi}{8}\right)=-(\sqrt{2}-1)=1-\sqrt{2}$

Given,

$\begin{aligned} & 1+\sec ^2 x \sin ^2 x \\ &=1+\left(\frac{1}{\cos ^2 x}\right) \sin ^2 x \\ &=1+\tan ^2 x=\sec ^2 x \end{aligned}$

$\begin{aligned} & \text { } \cos ^4 \theta=a \cos 4 \theta+b \cos 2 \theta+c \\ & \Rightarrow \cos ^4 \theta=a\left[2 \cos ^2 2 \theta-1\right]+b\left[2 \cos ^2 \theta-1\right]+c \\ & \Rightarrow \cos ^4 \theta=a \cdot 2 \cos ^2 2 \theta-a+2 b \cos ^2 \theta-b+c \\ & \Rightarrow \cos ^4 \theta=2 a(\cos 2 \theta)^2+2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=2 a\left(2 \cos ^2 \theta-1\right)^2+2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=2 a\left(4 \cos ^4 \theta+1-4 \cos ^2 \theta\right) \\ & +2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=8 a \cos ^4 \theta+2 a-8 a \cos ^2 \theta \\ & +2 b \cos ^2 \theta-a-b+c \\ & \Rightarrow \cos ^4 \theta=8 a \cos ^4 \theta+\cos ^2 \theta(-8 a+2 b)+a-b+c \\ \end{aligned}$

On comparing the corresponding coefficients, we get

$\begin{aligned} & 8 a=1,-8 a+2 b=0 \text { and } a-b+c=0 \\ & \Rightarrow \quad a=\frac{1}{8} \\ & \therefore \quad-8\left(\frac{1}{8}\right)+2 b=0 \text { gives } b=\frac{1}{2} \\ & \text { and } \quad \frac{1}{8}-\frac{1}{2}+c=0 \text { gives } c=\frac{3}{8} \text {. } \\ \end{aligned}$

$\begin{aligned} & \frac{\sin \theta+\sin 3 \theta}{\cos \theta+\cos 3 \theta} \\ & =\frac{\sin \theta+3 \sin \theta-4 \sin ^3 \theta}{\cos \theta+4 \cos ^3 \theta-3 \cos \theta} \\ & =\frac{4 \sin \theta-4 \sin ^3 \theta}{4 \cos ^3 \theta-2 \cos \theta} \\ & =\frac{4 \sin \theta \cdot\left(1-\sin ^2 \theta\right)}{2 \cos \theta\left(2 \cos ^2 \theta-1\right)} \\ & =\frac{4 \sin \theta \cdot \cos ^2 \theta}{2 \cos \theta \cdot \cos 2 \theta} \\ & =\frac{2 \sin \theta \cos \theta}{\cos 2 \theta} \\ & =\frac{\sin 2 \theta}{\cos 2 \theta}=\tan 2 \theta \end{aligned}$

$(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \ldots (1+\tan 45^{\circ})$

Considering the following two terms

$(1+\tan x)\left(1+\tan \left(45^{\circ}-x\right)\right)$

$\begin{aligned} & =1+\tan x+\tan \left(45^{\circ}-x\right)+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\left[\tan x+\tan \left(45^{\circ}-x\right)\right]+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\tan \left(x+\left(45^{\circ}-x\right)\right) \cdot\left[1-\tan x \cdot \tan \left(45^{\circ}-x\right)\right] \\ & \quad+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+\left[1-\tan x \cdot \tan \left(45^{\circ}-x\right)\right]+\tan x \cdot \tan \left(45^{\circ}-x\right) \\ & =1+1=2 \end{aligned}$

There are 22 such terms in the given multiplication

$\begin{aligned} & =(2)^{22} \cdot\left(1+\tan 45^{\circ}\right) \\ & =2^{22} \cdot 2^1=2^{23}=2^n \\ & \therefore \quad n=23 \end{aligned}$

$\begin{aligned} & \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta} \\ & =\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \\ & =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin ^2 \theta}{\sin \theta-\cos \theta} \\ & =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}-\frac{\sin ^2 \theta}{(\cos \theta-\sin \theta)} \\ & =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta-\sin \theta} \\ & =\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{(\cos \theta-\sin \theta)} \\ & =\cos \theta+\sin \theta \end{aligned}$

$\operatorname{cosech} x=\frac{4}{5}$

We know that, $\sinh x=\frac{1}{\operatorname{cosech} x}$

$\therefore \sinh x=\frac{5}{4}$

$\tan\theta=\frac{-5}{12}$

Then, $|\tan\theta|=5/12$

and $ - \tan \theta = - \left( {{{ - 5} \over {12}}} \right) = {5 \over {12}}$

$\therefore$ $|\tan \theta | = - \tan \theta $

$\begin{aligned} & \text { } \cos \frac{\pi}{4} \cos \frac{\pi}{8} \cos \frac{\pi}{16} \cos \frac{\pi}{32} \\ & =\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^5} \\ & =\frac{1}{2 \sin \left(\frac{\pi}{2^5}\right)} \\ & {\left[2 \sin \left(\frac{\pi}{2^5}\right) \cdot \cos \left(\frac{\pi}{2^5}\right) \cdot \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4}\right]} \\ & =\frac{1}{2} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^4} \cdot \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3}\right] \\ & =\frac{1}{4} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^2}\right] \\ & =\frac{1}{8} \operatorname{cosec}\left(\frac{\pi}{2^5}\right)\left[\sin \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^2}\right] \\ & =\frac{1}{16} \operatorname{cosec} \frac{\pi}{2^5} \sin \frac{\pi}{2} \\ & =2^{-4} \operatorname{cosec} \frac{\pi}{2^5} \quad\left[\because \sin \frac{\pi}{2}=1\right] \\ & \therefore m=-4 \text { and } n=2^5 \text {, then } \\ & m+n=-4+32=28 \\ & \end{aligned}$

$\begin{aligned} & A+B+C=\frac{3 \pi}{2} \\ & A+B=\frac{3 \pi}{2}-C \\ & \cos (A+B)=\cos \left(\frac{3 \pi}{2}-C\right)=-\sin C \end{aligned}$

Now, $\cos 2 A+\cos 2 B+\cos 2 C$

$\begin{aligned} & =2 \cos (A+B) \cos (A-B)+\cos 2 C \\ & =2(-\sin C) \cos (A-B)+\cos 2 C \\ & =-2 \sin C \cos (A-B)+1-2 \sin ^2 C \\ & =1-2 \sin C(\cos (A-B)+\sin C) \\ & =1-2 \sin C\left[\cos (A-B)+\sin \left(\frac{3 \pi}{2}-(A+B)\right)\right] \\ & =1-2 \sin C[\cos (A-B)+(-\cos (A+B)] \\ & =1-2 \sin C[\cos (A-B)-\cos (A+B)] \\ & =1-2 \sin C[2 \sin A \sin B] \\ & =1-4 \sin A \sin B \sin C \end{aligned}$

$\sinh (x+y) \cosh (x-y)=\frac{1}{2} [2 \sin \mathrm{h}(x+y) \cosh (x-y)]$ .... (i)

$\because \sin \mathrm{h} A+\sin \mathrm{h} B=2 \sin \mathrm{h}\left(\frac{A+B}{2}\right) \cosh \left(\frac{A-B}{2}\right)$

Let $\frac{A+B}{2}=x+y$

$\begin{aligned} & \Rightarrow A+B=2 x+2 y \\ & \text { and } A-B=2 x-2 y \end{aligned}$

Adding, we obtain $2 A=4 x$

$A=2 x$

Subtracting, we obtain $2 B=4 y$

$\begin{aligned} B & =2 y \\ \therefore 2 \sinh (x+y) \cosh (x-y) & =\sinh 2 x+\sin \mathrm{h} 2 y \\ \Rightarrow \sinh (x+y) \cosh (x-y) & =\frac{1}{2}[\sin \mathrm{h} 2 x+\sin \mathrm{h} 2 y] \end{aligned}$

$\begin{aligned} \cos \left(\frac{\pi}{8}\right) & =\sqrt{\frac{1+\cos \frac{\pi}{4}}{2}} \\ & =\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}=\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\end{aligned}$

$\begin{aligned} & \cos \theta=-\sqrt{\frac{3}{2}} \\ & \sin \alpha=-\frac{3}{5} \\ & \text { As, }-1 \leq \cos \theta \leq 1 \end{aligned}$

In the question given, $\cos \theta=-\sqrt{\frac{3}{2}}$. So, this is a wrong question.

$\begin{aligned} & \tan \beta=\frac{\tan \alpha+\tan \gamma}{1+\tan \alpha \tan \gamma} \\ & \tan \beta=\frac{\sin \alpha \cos \gamma+\sin \gamma \cos \alpha}{\cos \alpha \cos \gamma+\sin \alpha \sin \gamma} \\ & \tan \beta=\frac{\sin (\alpha+\gamma)}{\cos (\alpha-\gamma)} \\ & \frac{\sin 2 \alpha+\sin 2 \gamma}{1+\sin 2 \alpha \sin 2 \gamma} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+(2 \sin \alpha \cos \gamma)(2 \sin \gamma \cos \alpha)} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+[\sin (\alpha+\gamma)+\sin (\alpha-\gamma)][\sin (\alpha+\gamma)-\sin (\alpha-\gamma)]}\end{aligned}$

$\begin{aligned} & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+\sin ^2(\alpha+\gamma)-\sin ^2(\alpha-\gamma)} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{\sin ^2(\alpha+\gamma)+\cos ^2(\alpha-\gamma)}=\frac{2 \tan \beta}{1+\tan ^2 \beta}=\sin 2 \beta\end{aligned}$

$\begin{aligned} & \cos \left(\alpha+\frac{\pi}{2}\right)=-\sin \alpha \\ & \cos \left(\beta+\frac{\pi}{2}\right)=-\sin \beta \\ & \cos \left(\gamma+\frac{\pi}{2}\right)=-\sin \gamma \\ & \because \quad A M \geq G M \\ & \frac{-(\sin \alpha+\sin \beta+\sin \gamma)}{3} \geq(-\sin \alpha \sin \beta \sin \gamma)^{1 / 3} \end{aligned}$

Equating holds, when $\alpha=\beta=\gamma$

$\begin{gathered} \alpha+\beta+\gamma=360 \gamma \\ \alpha=\beta=\gamma=120 \gamma \\ \Rightarrow A M=\frac{1}{3}\left[\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{\sqrt{3}}{2}\right)\right] \\ =-\frac{\sqrt{3}}{2} \end{gathered}$

$3\sin A + 4\cos B = 6$ ...... (i)

and $4\sin B + 3\cos A = 1$ .... (ii)

Squaring Eqs. (i) and (ii), we get

$9{\sin ^2}A + 16{\cos ^2}B + 24\sin A\cos B = 36$ ..... (iii)

$16{\sin ^2}B + 9{\cos ^2}A + 24\sin B\cos A = 1$ .... (iv)

Adding Eqs. (iii) and (iv), we get

$\begin{aligned} & 9\left(\sin ^2 A+\cos ^2 A\right)+16\left(\sin ^2 B+\cos ^2 B\right) \\ & +24(\sin A \cos B+\cos A \sin B)=37 \\ & \Rightarrow \quad 25+24 \sin (A+B)=37 \\ & \sin (A+B)=1 / 2 \\ & \end{aligned}$

Let $\tan A+2 \cot 2 A$

$= \frac{\sin A}{\cos A}+\frac{2 \cos 2 A}{\sin 2 A}$

$= \frac{\sin A}{\cos A}+\frac{2\left(1-2 \sin ^2 A\right)}{2 \sin A \cos A}$

$= \frac{\sin ^2 A+1-2 \sin ^2 A}{\sin A \cos A}$

$= \frac{1-\sin ^2 A}{\sin A \cos A}=\frac{\cos ^2 A}{\sin A \cos A}=\cot A$

$\therefore \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha \\ = \tan \alpha+2 \tan 2 \alpha+4(\tan 4 \alpha+2 \cot 2(4 \alpha))$

$= \tan \alpha+2 \tan 2 \alpha+4 \cot 4 \alpha$

$= \tan \alpha+2(\tan 2 \alpha+2 \cot 2(2 \alpha))$

$= \tan \alpha+2 \cot 2 \alpha=\cot \alpha$

$\begin{aligned} & \text { If } f(x)=\frac{\cot x}{1+\cot x} \text { and } \alpha+\beta=\frac{5 \pi}{4} \\ & \Rightarrow \cot (\alpha+\beta)=\cot \frac{5 \pi}{4}=1 \Rightarrow \frac{\cot \alpha \cot \beta-1}{\cot \beta+\cot \alpha}=1 \\ & \Rightarrow \cot \alpha \cot \beta=\cot \beta+\cot \alpha+1\quad ....(\mathrm{i}) \\ & \text { Now, } f(\alpha) f(\beta)=\frac{\cot \alpha}{1+\cot \alpha} \times \frac{\cot \beta}{1+\cot \beta} \\ & =\frac{\cot \alpha \cot \beta}{1+\cot \alpha+\cot \beta+\cot \alpha \cot \beta}=\frac{\cot \alpha \cot \beta}{2 \cot \alpha \cot \beta}=\frac{1}{2}\end{aligned}$

In $\triangle A B C$,

$\begin{aligned} & A B=c, B C=a, A C=b \\ & \because \frac{a+b+c}{B C+A B}+\frac{a+b+c}{A C+A B}=3 \\ & \Rightarrow \frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3 \\ & \Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\ & \Rightarrow \quad \frac{b}{a+c}+\frac{a}{b+c}=1 \\ & \Rightarrow \quad \frac{a^2+b^2-c^2}{}=a b \\ & \Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=\frac{1}{2} \Rightarrow \cos C=\frac{1}{2} \\ & \Rightarrow \quad C=60 \Upsilon \end{aligned}$

Now, $\tan \frac{C}{8} =\tan \frac{60 \Upsilon}{8}$

$=\tan \left(7 \frac{1}{2}\right)^{\Upsilon}=(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)$

$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

To find the mean of the values given, $\sin^2 10Y, \sin^2 20Y, \sin^2 30Y, \ldots, \sin^2 90Y$, we first need to recognize a pattern or symmetry that can simplify our calculations.

Firstly, we note that the sine function has a property where $\sin^2$ of complementary angles add to 1. This is given by the identity:

$\sin^2 \theta + \cos^2 \theta = 1$

When dealing with complementary angles, we can write this as:

$\sin^2 (90^\circ - \theta) = \cos^2 \theta$

So, if we pair each element with its complement to $90^\circ$, we get:

$\sin^2 10Y = \cos^2 80Y$

$\sin^2 20Y = \cos^2 70Y$

$\sin^2 30Y = \cos^2 60Y$

$\sin^2 40Y = \cos^2 50Y$

$\sin^2 50Y = \cos^2 40Y$

$\sin^2 60Y = \cos^2 30Y$

$\sin^2 70Y = \cos^2 20Y$

$\sin^2 80Y = \cos^2 10Y$

$\sin^2 90Y = 1$

Essentially, the sum of each pair $\sin^2 \theta + \cos^2 \theta = 1$. Therefore, pairing these up (excluding $\sin^2 90Y$ as it stands alone giving 1), we see that the total number of pairs is 4 (since we started from 10 up to 80, which is 8 elements). Each pair sums to 1, so we have:

$\sin^2 10Y + \sin^2 80Y = 1$

$\sin^2 20Y + \sin^2 70Y = 1$

$\sin^2 30Y + \sin^2 60Y = 1$

$\sin^2 40Y + \sin^2 50Y = 1$

And $\sin^2 90Y = 1$

Thus, the total sum of the 9 terms is:

$4 \times 1 + 1 = 5$

Since there are 9 total terms, the mean is:

$\frac{\text{Total sum}}{\text{Number of terms}} = \frac{5}{9}$

Therefore, the correct answer is Option A: $\frac{5}{9}$

Let coordinate of $P$ original system be $(x, y)$.

Then, new coordinate after rotating axes through $135\Upsilon$ is $(4,-3)$

$\begin{aligned} & \therefore \quad 4=x \cos 135 x+y \sin 135\Upsilon \\ & \text { and }-3=-x \sin 135 \gamma+y \cos 135\Upsilon\\ & \Rightarrow \quad 4=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}} \\ & \text { and } \quad-3=\frac{-x}{\sqrt{2}}-\frac{y}{\sqrt{2}} \\ \end{aligned}$

On adding, $\quad 1=\frac{-2 x}{\sqrt{2}} \Rightarrow x=-\frac{1}{\sqrt{2}}$

On subtracting, $7=\frac{2 y}{\sqrt{2}} \Rightarrow y=\frac{7}{\sqrt{2}}$

Coordinate of $P$ in original system $\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

Given, $f(x)=\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Range of $\sin x=[-1,1]$

For $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \sin x$ is increasing function

It is maximum at $x=\frac{\pi}{2}$

$\operatorname{Max} f(x)=\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1$

$\begin{aligned} & \because \tan (A+B+C) \\ & \quad=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A} \end{aligned}$

If $A+B+C=90 \Upsilon$, then

$1-\tan A \tan B-\tan B \tan C-\tan C \tan A=0$

$\text { or } \tan A \tan B+\tan B \tan C+\tan C \tan A=1$

Here, $A=2 \alpha, B=30 \Upsilon-\alpha, C=60 \Upsilon-\alpha$, then

$\begin{aligned} & A+B+C=2 \alpha+30 Y-\alpha+60 Y-\alpha=90^{\circ} \\ & \therefore \tan 2 \alpha \tan (30 Y-\alpha)+\tan 2 \alpha \cdot \tan (60 Y-\alpha) \\ & +\tan (30 Y-\alpha) \tan (60 Y-\alpha)=1 \end{aligned}$

$\sin \alpha - \cos \alpha = m$ and $\sin 2\alpha = n - {m^2}$

$\therefore$ ${(\sin \alpha - \cos \alpha )^2} = {m^2}$

${\sin ^2}\alpha + {\cos ^2}\alpha - 2\sin \alpha \cos \alpha = {m^2}$

or $1 - \sin 2\alpha = {m^2}$

or $\sin 2\alpha = 1 - {m^2}$

But $\sin 2\alpha = n - {m^2}$ (given)

$ \Rightarrow n = 1$

$\begin{aligned} & \text { } \because \tan \theta=\sinh u \\ & \Rightarrow \quad \tan \theta=\sqrt{\cosh ^2 u-1} \quad\left[\cosh ^2 x-\sinh ^2 x=1\right] \\ & \Rightarrow \cosh u=\sqrt{1+\tan ^2 \theta} \Rightarrow \cosh u=\sec \theta\end{aligned}$