If tan theta_1 = k cot theta_2,then frac{cos | vedclass

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

Question

(B) Given that $	an 	heta_1 = k \cot 	heta_2$. Since $\cot 	heta_2 = \frac{1}{	an 	heta_2}$,we have $	an 	heta_1 = \frac{k}{	an 	heta_2}$,wh

Vedclass · Accepted Answer

$\frac{1-k}{1+k}$

Vedclass · Answer

(B) Given that $	an 	heta_1 = k \cot 	heta_2$. Since $\cot 	heta_2 = \frac{1}{	an 	heta_2}$,we have $	an 	heta_1 = \frac{k}{	an 	heta_2}$,which implies $	an 	heta_1 	an 	heta_2 = k$. Now,consider the expression $\frac{\cos (	heta_1 + 	heta_2)}{\cos (	heta_1 - 	heta_2)}$. Using the expansion formulas,we get $\frac{\cos 	heta_1 \cos 	heta_2 - \sin 	heta_1 \sin 	heta_2}{\cos 	heta_1 \cos 	heta_2 + \sin 	heta_1 \sin 	heta_2}$. Dividing the numerator and denominator by $\cos 	heta_1 \cos 	heta_2$,we obtain $\frac{1 - 	an 	heta_1 	an 	heta_2}{1 + 	an 	heta_1 	an 	heta_2}$. Substituting $	an 	heta_1 	an 	heta_2 = k$,the expression becomes $\frac{1-k}{1+k}$.

If $\tan \theta_1 = k \cot \theta_2$,then $\frac{\cos (\theta_1 + \theta_2)}{\cos (\theta_1 - \theta_2)} = $

Similar Questions

If $a \tan \alpha + b \tan \beta = (a + b) \tan \left( \frac{\alpha + \beta}{2} \right)$ and $\alpha - \beta \neq 2n\pi$,then $\frac{\cos \beta}{\cos \alpha} = $

The exact value of $\cos \frac{2\pi}{28} \csc \frac{3\pi}{28} + \cos \frac{6\pi}{28} \csc \frac{9\pi}{28} + \cos \frac{18\pi}{28} \csc \frac{27\pi}{28}$ is equal to

If $\cot x \cot y = a$ and $x+y = \frac{\pi}{6}$,then the quadratic equation whose roots are $\cot x$ and $\cot y$ is

The number of solutions of the equation $32^{\tan^{2} x} + 32^{\sec^{2} x} = 81$ for $0 \leq x \leq \frac{\pi}{4}$ is:

If $A$ and $B$ are acute angles satisfying $3 \cos^2 A + 2 \cos^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module