Question

(1/sinA-sinA)(1/cosA-cosA)=1/tanAcotA

Answer 1

Hey friend,

Here's your answer.

cosecA-sinA)(secA-cosA)
=(1/sinA – sinA )(1/cosA – cosA)
=[(1- sin2A)/sinA][(1-cos2A)/cosA]
=[(cos2A)/sinA][(sin2A)/cosA]
=sinA*cosA
\frac{sinA cosA}{sin^2A + cos^2A}

= \frac{1}{sin^2A/sinA cosA + cos^2A/sinA cosA }

=\frac{1}{tanA + cotA }
=rhs

I hope this helps you

Answer 2

Heyy Buddy ❤

Here's your Answer..

(1/sinA-sinA)(1/cosA-cosA)=1/tanAcotA

L.H.S.

=>[ (1 - sin^2A)/sinA ] [ (1 - cos^2A)/cosA]

=> (Cos^2A/ SinA ) ( Sin^2A/ CosA)

=> (SinA × CosA) /1

we know that, Sin^2A + Cos^2A = 1

So,

=> (SinA × CosA) /( sin^2A + Cos^2A)

=> Dividing Denominator and Numerator by SinA × CosA

=>[(siA× cosA)/ (sinA × CosA)]/ [(sin^2A / sinA × CosA) + Cos^2A/ sinA × CosA]

=> 1/ [(sin^2A / sinA × CosA) + Cos^2A/ sinA × CosA]

=> 1 / [(sinA/CosA) + (CosA/ sinA)]

=> 1 / [ tan A + CotA]

=> 1/ Tan A + cotA.
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