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Question
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Solution
L.H.S =(1 + cot θ – cosec θ)(1+ tan θ + sec θ)
= `(1 + costheta/sintheta - 1/sin theta)(1+sin theta/cos theta + 1/cos theta)`
`= ((sintheta + costheta - 1)/sintheta)((costheta + sintheta +1)/costheta)`
`= 1/(sinthetacostheta) ((sinthetacostheta+sin^2theta + sin theta+cos^2theta),(+sinthetacostheta+costheta-costheta - sin theta -1))`
`= 1/(sinthetacostheta) (2sinthetacostheta + (sin^2theta + cos^2 theta) - 1)`
`= 1/(sin thetacostheta) (2sinthetacostheta + 1 - 1)`
`= (2sin thetacostheta)/(sin thetacos theta)`
= 2
= R.H.S
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