Prove - 1- 2 - 1-tan 2 - - 1- - 1-tan - 2

LHS=1+^2θ1+tan^2θ=cosec^2θsec^2θ=^2θ RHS=(1+θ1+tanθ)^2=(sinθ+cosθsinθsinθ+cosθcosθ)^2=^2θ LHS=RHS Hence Proved ...

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Sum of Trigonometric Ratios in Terms of Their Product

Prove : 1+ ...

Question

Prove : $\frac{1 + {cot}^{2} θ}{1 + {tan}^{2} θ} = {(\frac{1 + {cot} θ}{1 + {tan} θ})}^{2}$

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\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2} = {tan}^{2} θ

\frac{sec θ - 1}{sec θ + 1} = {tan}^{2} \frac{θ}{2}

\frac{tan θ}{(1 + {tan}^{2} θ)^{2}} + \frac{cot θ}{(1 + {cot}^{2} θ)^{2}} = sin θ \cdot cos θ

θ

[\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ}] = [{\frac{1 - tan θ}{1 - cot θ}]}^{2} = {tan}^{2} A

{cos}^{2} θ = \frac{1}{1 + {tan}^{2} θ}

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