The value of frac{3 + cot 76^circ cot 16^circ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given expression: $E = \frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$.Converting to sine and cosine: $E = \frac{3 + \frac{

Vedclass · Accepted Answer

$\cot 44^\circ$

Vedclass · Answer

(A) Given expression: $E = \frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$.Converting to sine and cosine: $E = \frac{3 + \frac{\cos 76^\circ \cos 16^\circ}{\sin 76^\circ \sin 16^\circ}}{\frac{\cos 76^\circ}{\sin 76^\circ} + \frac{\cos 16^\circ}{\sin 16^\circ}} = \frac{3 \sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ}{\cos 76^\circ \sin 16^\circ + \sin 76^\circ \cos 16^\circ}$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,the denominator becomes $\sin(76^\circ + 16^\circ) = \sin 92^\circ$.For the numerator,write $3 = 2 + 1$: $2 \sin 76^\circ \sin 16^\circ + (\sin 76^\circ \sin 16^\circ + \cos 76^\circ \cos 16^\circ)$.Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$,the term in brackets is $\cos(76^\circ - 16^\circ) = \cos 60^\circ = \frac{1}{2}$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,the first term is $\cos 60^\circ - \cos 92^\circ = \frac{1}{2} - \cos 92^\circ$.Numerator $= \frac{1}{2} - \cos 92^\circ + \frac{1}{2} = 1 - \cos 92^\circ$.Thus,$E = \frac{1 - \cos 92^\circ}{\sin 92^\circ} = \frac{2 \sin^2 46^\circ}{2 \sin 46^\circ \cos 46^\circ} = 	an 46^\circ = \cot(90^\circ - 46^\circ) = \cot 44^\circ$.

The value of $\frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$ is:

Similar Questions

The value of $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$ is

If $90^{\circ} < \alpha < 180^{\circ}$,$\sin \alpha = \frac{\sqrt{3}}{2}$ and $180^{\circ} < \beta < 270^{\circ}$,$\sin \beta = -\frac{\sqrt{3}}{2}$,then $\frac{4 \sin \alpha - 3 \tan \beta}{\tan \alpha + \sin \beta} = $

$1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ = $

If $\tan \theta = \sqrt{\frac{3}{2}}$,the sum of the infinite series $1 + 2(1 - \cos \theta) + 3(1 - \cos \theta)^2 + 4(1 - \cos \theta)^3 + \dots \infty$ is

Let $f(x) = \sin^2 x + \cos^4 x + 2$ and $g(x) = \cos(\cos x) + \cos(\sin x)$. If the periods of $f(x)$ and $g(x)$ are $T_1$ and $T_2$ respectively,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module