Question

) Prove that:
COS X
sinx
+
= COS X + sin x
1-tan x 1-cotx
11.​

Answer 1

Answer:

L.H.S. = \dfrac{\sf cos\:x}{\sf 1 - tan\:x} + \dfrac{\sf sin\:x}{\sf 1 - cot\:x}1−tanxcosx+1−cotxsinx 

= \dfrac{\sf cos^{2}x}{\sf cos\:x(1 - tan\:x)} + \dfrac{\sf sin^{2}x}{\sf sin\:x(1 - cot\: x)}cosx(1−tanx)cos2x+sinx(1−cotx)sin2x 

= \dfrac{\sf cos^{2}x}{\sf cos\:x - sin\:x} + \dfrac{\sf sin^{2}x}{\sf sin\:x - cos\:x}cosx−sinxcos2x+sinx−cosxsin2x 

= \dfrac{\sf cos^{2}x}{\sf cos\:x - sin\:x} - \dfrac{\sf sin^{2}x}{\sf cos\:x - sin\:x}cosx−sinxcos2x−cosx−sinxsin2x 

= \dfrac{\sf cos^{2}x - sin^{2}x}{\sf cos\:x - sin\:x}cosx−sinxcos2x−sin2x 

= \dfrac{\sf (cos\:x + sin\:x) \cancel{(cos\:x - sin\:x)}}{\sf \cancel{cos\:x - sin\:x}}cosx−sinx(cosx+sinx)(cosx−sinx) 

= \sf cos\:x + sin\:xcosx+sinx = R.H.S.

\boxed{\sf

HENCE\: PROVED}HENCEPROVED