Question

1/1+cos theta +1/1-cos theta=2cosec square theata

Answer 1

Hello there !

$\dfrac{1}{1+cosA} + \dfrac{1}{1-cosA}$

$\dfrac{1(1-cosA)+1(1+cosA)}{(1+cosA)(1-cosA)}$

We know that ( a +  b )( a - b ) is equal to a^2 - b^2. In the question, on denominator the same condition is arising. ∴ ( 1 - cosA )( 1 + cosA ) = 1^2 - cos^2 A . As we know, square of 1 [ 1^2 ] is always 1. So, ( 1 - cosA )( 1 + cosA ) = 1 - cos^2 A .

$\dfrac{1+1+cosA-cosA}{1-cos^{2}A}$

From trigonometric identities, 1 - cos^2 A = sin^2 A

$\dfrac{2}{sin^{2}A }$

$2\times \dfrac{1}{sin^{2}A}$

From trigonometric identities, 1 / sin^2 A = cosec^2 A

$2 \times cosec^{2}A$

Hence, proved that 1 / 1 + cosA + 1 / 1 - cosA = 2 cosec^2 A

Answer 2

Step-by-step explanation:

hope this helps you....