tan left( frac{pi }{4} + theta right) - tan l | vedclass

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

Question

(A) We use the expansion formulas $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.

Vedclass · Accepted Answer

$2	an 2	heta $

Vedclass · Answer

(A) We use the expansion formulas $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Given expression: $	an \left( \frac{\pi }{4} + 	heta ight) - 	an \left( \frac{\pi }{4} - 	heta ight)$.Substituting $	an \frac{\pi }{4} = 1$,we get:$= \frac{1 + 	an 	heta}{1 - 	an 	heta} - \frac{1 - 	an 	heta}{1 + 	an 	heta}$.Taking the common denominator:$= \frac{(1 + 	an 	heta)^2 - (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$.Using $(a+b)^2 - (a-b)^2 = 4ab$ and $(a-b)(a+b) = a^2 - b^2$:$= \frac{4 	an 	heta}{1 - 	an^2 	heta}$.$= 2 \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight)$.Since $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,the expression simplifies to:$= 2 	an 2	heta$.

$\tan \left( \frac{\pi }{4} + \theta \right) - \tan \left( \frac{\pi }{4} - \theta \right) = $

Similar Questions

$\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4} \cdots \cos \frac{\pi}{2^{10}} = $

$\sqrt{2+\sqrt{2+2 \cos 4 \theta}} = $

$A$ true statement among the following identities is

$\cot \theta - \tan \theta - 2 \tan 2 \theta - 4 \tan 4 \theta = $

If $x \in \left(0, \frac{\pi}{2}\right)$ and $x$ satisfies the equation $\sin x \cos x = \frac{1}{4}$,then the values of $x$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module