Prove the following identity, where the angle involved is acute angle for which the expressions are defined. (cosA-sinA+1)/cosA+sinA-1)=cosecA+cotA.

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asked Feb 5, 2018 in Mathematics by Kundan kumar (49,132 points) 34 441 1404

Prove the following identity, where the angle involved is acute angle for which the expressions are defined.

(cosA-sinA+1)/(cosA+sinA-1)=cosecA+cotA.

using the identity cosec2A=1+cot2A.
 

2 Answers

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answered Feb 5, 2018 by Vikash Kumar (144,729 points) 8 11 26
 
Best answer

Answer is....

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answered Feb 6, 2018 by pawan (15 points)
cosA-sinA+1/cosA +sinA-1 = cosecA+cotA

Use the formula cot^2A-cosec^2A

LHS-   cosA-sinA+1/cosA+sinA-1

divide by sinA on both numerator and denominator

cosA/sinA-sinA/sinA+1/sinA /cosA/sinA-sinA/sinA+1/sinA

cotA-1+cosecA/cotA+1-cosecA

at the place of 1 put above formula

cotA+cosec-(cot^2A-cosec^2A)/cotA-cosecA+1

now put it in a^2-b^2 =(a+b)(a-b)

cotA+cosecA+(cotA-cosecA)(cotA+cosecA)/cotA-cosecA+1

now common taking from numerator

cotA+cosecA(1+cotA-cosecA)/cotA-cosecA+1

Hence it prove cotA+cosecA

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