Question

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Answer 1

Answer:

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Answer 2

[tex]\mathfrak{\huge{Answer:-}} \\ \bold{ \frac{1 + sinθ - cosθ}{cosθ} \times \frac{1 - sinθ + cosθ}{cosθ} = 2tanθ} \\ \frac{1 + (sinθ - cosθ)}{cosθ} \times \frac{1 - ( sinθ - cosθ)}{cosθ} = 2tanθ \\ \frac{ {(1)}^{2} - {(sinθ - cosθ)}^{2} }{ {cos}^{2}θ } = 2tanθ \\ \frac{1 - ( {sin}^{2}θ + {cos}^{2} θ - 2sinθcos θ)}{ {cos}^{2}θ } \\ \frac{1 - (1 - 2sinθcosθ)}{ {cos}^{2}θ } = 2tanθ \\ \frac{1 - 1 + 2sinθcosθ}{ {cos}^{2} θ} = 2tanθ \\ \frac{2sinθcosθ}{ {cos}^{2}θ } = 2tanθ \\ \frac{2sinθ}{cosθ} = 2tanθ \\ 2 \times \frac{sinθ}{cosθ} = 2tanθ \\ 2tanθ = 2tanθ \\ LHS=RHS \\ Hence \: proved[/tex]