1. ## Trigonometry identity proof

prove that

(tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1

2. Originally Posted by saha.subham
prove that

(tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1
$\displaystyle 1 + \tan ^2 a = \sec ^2 a$

$\displaystyle 1 + \cot ^2 a = \csc ^2 a$

$\displaystyle \sec x = \frac{1}{\cos x}$

$\displaystyle \csc x = \frac{1}{\sin x}$

$\displaystyle \sin ^2 x + \cos ^2 x = 1$

$\displaystyle \frac{\tan ^2 \theta }{\sec ^2 \theta } + \frac{\cos ^2 \theta }{\csc ^2 \theta }$

......continue it

3. plzz elaborate it , not getting it. i am new to this chapter

4. Originally Posted by saha.subham
prove that

(tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1
$\displaystyle \frac{\tan ^2 \theta}{1+\tan ^2 \theta} + \frac{\cot ^2 \theta}{1+\cos ^2 \theta}$

$\displaystyle \tan x = \frac{\sin x}{\cos x}$

$\displaystyle \cot x = \frac{\cos x}{\sin x}$

sub tan and cot values

$\displaystyle \frac{\frac{\sin ^2 x}{\cos ^2 x}}{1 + \frac{\sin ^2 x}{\cos ^2 x}}+\frac{\frac{\cos ^2 x}{\sin ^2 x}}{1+\frac{\cos ^2 x}{\sin ^2 x}}$

in the first fraction multiply both the denominator and nominator by $\displaystyle \cos ^2 x$ and the second fraction by $\displaystyle \sin ^2 x$

you will have this

$\displaystyle \frac{\sin ^2 x}{\sin ^2 x+ \cos ^2 x } + \frac{\cos ^2 x}{\cos ^2 x +\sin ^2 x}$

but $\displaystyle \sin ^2 x + \cos ^2 x = 1$
so

$\displaystyle \frac{\sin ^2 x }{1} + \frac{\cos ^2 x }{1} = \sin ^2 x + \cos ^2 x = 1$ ends

5. Hello, saha.subham!

A slightly different approach . . .

Prove that: .$\displaystyle \frac{\tan^2\!\theta}{1 + \tan^2\!\theta} + \frac{\cot^2\!\theta}{1 + \cot^2\!\theta} \;=\;1$

We have: .$\displaystyle \frac{\tan^2\!\theta}{1+\tan^2\!\theta} + \frac{\cot^2\!\theta}{1 + \cot^2\!\theta} \;\;=\;\; \frac{\tan^2\!\theta}{\sec^2\!\theta} + \frac{\cot^2\!\theta}{\csc^2\!\theta}$ . $\displaystyle =\;\;\frac{\dfrac{\sin^2\!\theta}{\cos^2\!\theta}} {\dfrac{1}{\cos^2\!\theta}} + \frac{\dfrac{\cos^2\!\theta}{\sin^2\!\theta}}{\dfr ac{1}{\sin^2\!\theta}} \;\;=\;\;\sin^2\!\theta + \cos^2\!\theta \;\;=\;\;1$