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Question
Prove the following :
`(cos(90^@ - theta) sec(90^@ - theta)tan theta)/(cosec(90^@ - theta) sin(90^@ - theta) cot (90^@ - theta)) + tan (90^@ - theta)/cot theta = 2`
Solution
Cos (90° - θ) = sin A cosec (90 - θ) = sec θ
Sec (90° - θ) = cosec θ sin (90 - θ) = cos θ
Cot (90 - θ) = tan θ
`=> (sin theta cosec theta tan theta)/(sec theta. cos theta. tan theta) = (sin theta cosec theta)/(sec theta cos theta)` ` [∵ sin theta cosec theta = 1]`
=1 `[sec theta cos theta = 1]`
`tan (90^@ - theta)/cot theta = cot theta/cot theta = 1`
=> 1 + 1= 2
`:. LHS = RHS`
Hence proved
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