Prove that 1-tan 2 - 1- 2 - - 1-tan- 1- 2 if - lies in the first quadrant-

LHS=1+tan^2θ1+^2θ=sec^2θcosec^2θ=tan^2θ RHS=(1 tanθ1 θ)^2=(1 tanθ1 1tanθ)^2=(1 tanθ1 tanθtanθ)^2=tan^2θ LHS=RHS hence proved ...

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General Solution of Trigonometric Equation

Prove that ...

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Prove that $\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2}$ if $θ$ lies in the first quadrant.

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

θ

c o s θ = \frac{8}{17},

c o s (\frac{π}{6} + θ) + c o s (\frac{π}{4} - θ) + c o s (\frac{2 π}{3} - θ)

= (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) \frac{23}{17}

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

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

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

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