Question

Prove the following trigonometric identities:
(1+cotA+tanA)(sinA-cosA)=secA/cosec²A-cosecA/sec²A=sinAtanA-cotAcosA

Answer 1

we have to prove that,

(1 + cotA + tanA)(sinA - cosA) = secA/cosec²A -cosecA/sec²A = sinA.tanA - cotA.cosA

secA/cosec²A - cosecA/sec²A

= (1/cosA)/(1/sin²A) - (1/sinA)/(1/cos²A)

= sin²A/cosA - cos²A/sinA

= (sin³A - cos³A)/sinA.cosA

= (sinA - cosA)(sin²A + cos²A + sinA.cosA)/sinA.cosA

= (sinA - cosA)[sin²A/sinA.cosA + cos²A/sinA.cosA + sinA.cosA/sinA.cosA]

= (sinA - cosA)(tanA +cotA + 1) ...........(1)

again, secA/cosec²A - cosecA/sec²A

= sin²A/cosA - cos²A/sinA

= (sinA/cosA)sinA - (cosA/sinA)cosA

= tanA.sinA - cotA.cosA ........(2)

from equations (1) and (2) we get,

(1 + cotA + tanA)(sinA - cosA) = secA/cosec²A -cosecA/sec²A = sinA.tanA - cotA.cosA

[ hence proved ]

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Answer 2

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