If y = (1 + tan A)(1 - tan B) where A - B = f | vedclass

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

Question

(C) Given $A - B = \frac{\pi}{4}$.Taking tangent on both sides,$	an(A - B) = 	an\frac{\pi}{4} = 1$.Using the formula $	an(A - B) = \frac{	an A - \

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) Given $A - B = \frac{\pi}{4}$.Taking tangent on both sides,$	an(A - B) = 	an\frac{\pi}{4} = 1$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = 1$.This implies $	an A - 	an B = 1 + 	an A 	an B$.Rearranging,$	an A - 	an B - 	an A 	an B = 1$.Adding $1$ to both sides,$	an A - 	an B - 	an A 	an B + 1 = 1 + 1 = 2$.Factoring the expression,$(1 + 	an A)(1 - 	an B) = 2$.Since $y = (1 + 	an A)(1 - 	an B)$,we have $y = 2$.Therefore,$(y + 1)^{y + 1} = (2 + 1)^{2 + 1} = 3^3 = 27$.

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi}{4}$,then $(y + 1)^{y + 1}$ is equal to

Similar Questions

If $3 \sin \theta + 4 \cos \theta = 3$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\sin 2 \theta = $

If $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$,then the value of the expression $\frac{\sin(A + B)}{\sin 2C}$ is:

Let $A = \{x \in R : |\sqrt{3} \cos x - \sin x| \geq 2, 0 \leq x \leq 2\pi\}$. If $x_1 \in A$ and $x_2 \in A$,then find the possible value of $\frac{x_1}{x_2}$.

Let $E = \left( 1 - \frac{\cos 61^\circ}{\cos 1^\circ} \right) \left( 1 - \frac{\cos 62^\circ}{\cos 2^\circ} \right) \dots \left( 1 - \frac{\cos 119^\circ}{\cos 59^\circ} \right)$,then $E$ is equal to:

If $\alpha, \beta$ are acute angles such that $\frac{\sin \alpha}{\sin \beta} = \frac{6}{5}$ and $\frac{\cos \alpha}{\cos \beta} = \frac{9}{5 \sqrt{5}}$,then $\sin \alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module