Question

Prove that
Sin^2(x)tan(x)+cos^2(x)cot(x)+2sin(x)cos(x)=(sin(x)+cos(x))/sin(x)cos(x)

Answer 1

\[\large{\sin^2 x \tan x + \cos^2 x \cot x + 2\sin x \cos x = \cfrac{(\sin x + \cos x)}{\sin x \cos x}}\]

Okay this is the question right ?

Answer 2

Yes

Answer 3

Write tan(x) and cot(x) in terms of sin(x) and cos(x).
Find the common denominator for the LHS.
Add the terms and simplify.

Answer 4

First of all:

\[\large{\tan x = \cfrac{\sin x}{\cos x}}\]

\[\large{\cot x = \cfrac{1}{\tan x} = \cfrac{\cos x}{\sin x}}\]

Use these in the LHS of the identity to be proved

Answer 5

Can you go a bit further

Answer 6

@Swqaedf You should attempt to use this information he has so helpfully given to see how far you can get! Good luck!

Answer 7

I tried this but I end up with LHS being \[ \frac{1}{\sin x \cos x}\]
I tried showing .. which seems to agree with Wolfram which simplified it to \(\csc x \sec x\): http://www.wolframalpha.com/input/?i=sin%5E2x+tan+x+%2B+cos%5E2+x+cot+x+%2B+2+sin+x+cos+x

Answer 8

Okay let me tell you the first few steps:

\[\large{\sin^2 (x) \tan (x)+\cos^2 (x) \cot (x)+2\sin (x) \cos (x)}\]
\[\large{=\sin^2 x \cfrac{\sin x}{\cos x} + \cos^2 x \cfrac{\cos x}{\sin x} + 2\sin x \cos x}\]

\[\large{=\cfrac{\sin^3 x}{\cos x} + \cfrac{\cos^3 x}{\sin x} + 2\sin x \cos x}\]

\[\large{= \cfrac{\sin^4 x + \cos^4x + 2\sin^2 x \cos^2 x}{\sin x \cos x}}\]

Now we use:

\[\large{a^4+b^4+2a^2b^2 = (a^2+b^2)^2}\]

Thus the LHS will become:

\[\large{=\cfrac{(\sin^2 x + \cos^2 x)^2}{\sin x \cos x}}\]

Now you should know that:

\[\large{\sin^2 x+ \cos^2 x = 1}\]

Use this to simplify the LHS and I tend to agree with @kirbykirby . There appears to be a mistake in the question. The RHS should be:

\[\large{\cfrac{1}{\sin x \cos x}}\]