Question

Important question for 12th .

Inverse trigonometry . ​

Answer 1

Hii

Step-by-step explanation:

#Answerwithquality#BAL

Solution is in this attachment .

Hope it's helpful

@Raj ✌

Answer 2

Answer:

2b/a

Step-by-step explanation:

Given,

$tan(\frac{\pi }{4} + \frac{1}{2} cos^{-1} \frac{a}{b}) + tan(\frac{\pi }{4} - \frac{1}{2} cos^{-1} \frac{a}{b})$

Let cos⁻¹(a/b) = ∅,

Then, cos∅ = (a/b) and (1/2) cos⁻¹(a/b) = ∅/2.

Now,

$\Rightarrow tan[\frac{\pi }{4} + \frac{\theta}{2}] +  tan[\frac{\pi }{4} - \frac{\theta}{2}]$

$\Rightarrow \frac{tan\frac{\pi }{4} + tan\frac{\theta}{2} }{1 - tan\frac{\pi }{4} tan\frac{\theta}{2}} + \frac{tan\frac{\pi }{4} - tan\frac{\theta}{2} }{1 + tan\frac{\pi }{4} tan\frac{\theta}{2}}$

$\Rightarrow \frac{1 + tan\frac{\theta}{2}}{1 - tan\frac{\theta}{2}} + \frac{1 - tan\frac{\theta}{2}}{1 + tan\frac{\theta}{2}}$

$\Rightarrow \frac{(1 + tan\frac{\theta}{2})^2 + (1 - tan\frac{\theta}{2})^2}{1 - tan^2\frac{\theta}{2} }$

$\Rightarrow 2(\frac{1}{cos\theta})$

$\Rightarrow \frac{2b}{a}$

Hope it helps!