Prove:tan θ + 1/ tan θ = sec θ cosec θ

1.	Prove:tan θ + 1/ tan θ = sec θ cosec θ
Answer» We have, L.H.S = tan θ + \(\frac{1}{tan\, θ}\) = \(\frac{(tan^2 θ\, +\, 1)}{tan θ}\) = \(\frac{sec^2 \,θ }{tan \,θ}\) [∵ sec² θ − tan² θ = 1] = (\(\frac{1}{cos^2\, θ}\)) x \(\frac{1}{sinθ \over cos θ}\) [∵ tan θ = \(\frac{sin \,θ }{cos \,θ}\)] = \(\frac{cos \,θ}{(sin θ \,\times\, cos^2\, θ)}\) = \(\frac{1}{cos\, θ}\) x \(\frac{1}{sin\, θ}\) = sec θ x cosec θ = sec θ cosec θ = R.H.S Hence Proved

Answer»

We have,

L.H.S = tan θ + \(\frac{1}{tan\, θ}\)

= \(\frac{(tan^2 θ\, +\, 1)}{tan θ}\)

= \(\frac{sec^2 \,θ }{tan \,θ}\) [∵ sec² θ − tan² θ = 1]

= (\(\frac{1}{cos^2\, θ}\)) x \(\frac{1}{sinθ \over cos θ}\) [∵ tan θ = \(\frac{sin \,θ }{cos \,θ}\)]

= \(\frac{cos \,θ}{(sin θ \,\times\, cos^2\, θ)}\)

= \(\frac{1}{cos\, θ}\) x \(\frac{1}{sin\, θ}\)

= sec θ x cosec θ

= sec θ cosec θ

= R.H.S

Hence Proved

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