Question

(cosecA-sinA)(SecA-cosA)=1/tana+cotA

plz solve tom its urgent yrr ​

Answer 1

Step-by-step explanation:

In left hand side

(cosecA-sinA)(secA-cosA)

(1/sinA-sinA)(1/cosA-cosA)

(1-sinA^2/sinA)(1-cos^2/cosA)

(cosA^2/sinA)(sinA^2/cosA)

SinAcosA......are left in left hand side

Now...solve right hand side

1/tanA+cotA

1/(sinA/cos+cos/sin)

1/(sin^2+cos^2/sincos)

sinAcosA...are left in right hand side...

So...

sinAcosA=sinAcosA

lhs=rhs...

:)Hope it will help u

..:)) Mark as brainlist answer plzzz.

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Answer 2

Answer:

refer to the above attachment