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`

Answer 1

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