Question

Solve it please.....

Answer 1

Hey !!!

From LHS

1 + cos¢ - sin²¢
----------------------
sin¢ (1 + cos¢ )

{Rearranging term }

1 - sin²¢ + cos¢
----------------------
sin¢ (1 + cos¢ )

[ •°• 1 - sin²¢ = cos²¢ ]

cos²¢ + cos¢
------------------
sin¢ ( 1 + cos¢ )

cos¢ ( cos¢ + 1 )
-----------------------[Taking common ]
sin¢ (cos¢ + 1 )

cos¢ /sin¢ = cot¢ RHS prooved

________________________
Hope it helps you !!!

@Rajukumar111

Answer 2

Here I am writing theta as A.

$\frac{1+cosA-sin^2A}{sina(1+cosa)}$

We know that 1 - sin^2a = cos^2a

$= \ \textgreater \ \frac{cosa+cos^2a}{sina(1+cosa)}$

$= \ \textgreater \ \frac{cosa(cosa+1)}{sina(cosa+1)}$

$= \ \textgreater \ \frac{cosa}{sina}$

cota.

Hope this helps!