Question

Prove that the given equations are identities:

a. (1+tan^2x)(1+cos2x)=2
b. (cosxcscx)/(tanx+cotx)=sin^2x+cos2x

Please help with one or both! I've gotten both part way but am messing up somewhere...

Answer 1

\(\cos 2x = \cos^2 x - \sin^2 x\)
\(1+ \tan^2 x = \sec ^2 x\)

Use the above formulas to prove i)

Answer 2

Here is how to do it :
\(\mathsf{(1+\tan^2 x)(1+\cos 2x) \\

(\sec^2 x)( 1+ \cos^2 x - \sin^2 x)
(\sec^2 x)(1 - \sin^2 x + \cos^2 x) \\
(\sec^2 x)(\cos^2 x + \cos^2 x) \\
(\sec^2 x)(2\cos^2 x)\\
}\)

Write : \(\sec^2 x = \cfrac{1}{\cos^2 x}\)

Answer 3

prove it then.

Answer 4

For ii)

\(\mathsf{(\cos x \csc x)/(\tan x+\cot x)=\sin^2 x+\cos 2x}\)

Hint : Write csc x = 1/sinx

tanx = sinx/cos x

cot x = cos x / sin x

Answer 5

And for the second, I had originally gotten cotx/(tanx+cotx)=cos^2x from using the double angle formula and combining the numerator, but I wasn't sure where to go from there

Answer 6

\(\cfrac{\cfrac{1}{\tan x} }{\tan x + \cfrac{1}{\tan x} } = \cos^2x\)

Answer 7

\(\cfrac{\cfrac{1}{\tan x} }{\cfrac{\tan^2 x + 1}{\tan x}} = \cos^2 x\)

\(\cfrac{1}{\sec^2 x} = \cos^2 x \)

Answer 8

This is what I did :

\(\cfrac{\cfrac{1}{\tan x} . \tan x }{\tan^2 x + 1} \)

\(\cfrac{1}{\tan^2 x + 1}\)

\(\tan^2 x + 1 = \sec^2 x\) -- identitty

\(\cfrac{1}{\sec^2 x} = \cos^2 x \cdot \bf{Proved!} \)

Answer 9

@elly394 , got it?

Answer 10

Yes I did!! Thank you soooo much!!

Answer 11

You're welcome :)