Question

tan(45+x)-tan(45-x)=2sec2x

Answer 1

I think the question is wrong.....there should be 2sex2xsin2x on the RHS
the proof is given in the image attached...please mark as brainliest if you find it useful....

Answer 2

Answer:

Given: $\tan (45+x)+\tan (45-x)=2 \sec 2 x$

Consider L H S, multiplying the values in the bracket,

$\begin{array}{l}{=\frac{\tan 45+\tan x}{1-\tan 45 \tan x}+\frac{\tan 45-\tan x}{1+\tan 45 \tan x}} \\ {=\frac{1+\tan x}{1-\tan 45 \tan x}+\frac{1-\tan x}{1+\tan 45 \tan x}} \\ {=\frac{\cos x+\sin x}{\cos x-\sin x}+\frac{\cos x-\sin x}{\cos x+\sin x}} \\ {=\frac{(\cos x+\sin x)(\cos x+\sin x)}{\cos 2 x-\sin 2 x}} \\ {=\frac{2}{\cos 2 x}} \\ {=2 \sec 2 x}\end{array}$

Therefore, $\tan (45+x)+\tan (45-x)=2 \sec 2 x$

Hence proved

Step-by-step explanation: