Question

Prove that

tan θ tan θ + sec θ + 1
_____ = ______________
sec θ tan θ + sec θ - 1​

Answer 1

$\huge\sf\purple{\underline{\underline{Correct \: Question :-}}}$

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$\sf \dfrac{tan \: θ}{sec \: θ - 1} \: = \: \sf \dfrac{tan \: θ + sec \: θ + 1}{tan \: θ + sec \: θ - 1}$

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$\huge\sf{\underline{\underline{Solution :-}}}$

$\sf{}$

Proof :

$\implies\tt{1 + {tan}^{2} \: θ} = {{sec}^{2}} \: θ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \tt{ {tan}^{2} \: θ = \: {sec}^{2} \: θ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\tt{tan \:θ \: \: tan \: θ =( sec \:θ + 1) \: (sec \: θ - 1) } \\ \\ \implies\tt \dfrac{tan \: θ}{sec \:θ \: - 1 } \: = \: \tt\dfrac{sec \: θ + 1}{tan \: θ} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\implies\tt\dfrac {tan \:θ }{sec \: θ - 1} \: = \: \dfrac{tan \:θ + sec \: θ + 1}{sec \: θ - 1 + tan \: θ} \: \: .....(Theorem \: on \: equal \: ratios)$$\sf{}$

$\implies \tt \dfrac{tan \: θ}{sec \: θ - 1} \: = \: \tt \dfrac{tan \: θ + sec \: θ + 1}{tan \:θ + sec \: θ - 1 }$

Answer 2

Refer the attachment......Pag1 is the answer of ur question pag2 is additional prooof to other identity