Question

29. Prove that, (cosec -cot)^2 = 1 - cos÷
1 + cos ​

Answer 1

Answer:

( cosecA - cotA )^2 = ( 1 - cosA ) / ( 1 + cosA )

Step-by-step explanation:

= > ( cosecA - cotA )^2

From the properties of trigonometric functions :

• cosecB = 1 / sinA
• cotB = cosB / sinB

= > { ( 1 / sinA ) - ( cosA / sinA ) }^2

= > ( 1 / sin^2 A ) { 1 - cosA }^2

= > ( 1 - cosA )^2 / sin^2 A

= > { ( 1 - cosA )( 1 - cosA ) } / { 1 - cos^2 A } { sin^2 A = 1 - cos^2 A }

= > { ( 1 - cosA )( 1 - cosA ) } / { ( 1 + cosA )( 1 - cosA ) } { using a^2 - b^2 = ( a + b )( a - b ), 1 - cos^2 A = ( 1 + cosA )( 1 - cosA ) }

= > ( 1 - cosA ) / ( 1 + cosA ) { 1 - cosA was common in numerator & denominator, which has been cancelled out }

Hence, ( cosecA - cotA )^2 = ( 1 - cosA ) / ( 1 + cosA )