Question

prove that 1 / 1-sin + 1 / 1+sin = 2sec^2 ​

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:\frac{1}{1-Sin θ} + \frac{1}{1+Sin θ} = 2 Sec^2 θ$

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$\large\sf\underline{To\::}$

• Prove LHS = RHS

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$\large\sf\underline{Solution\::}$

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• Taking LHS

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$\sf\:\frac{1}{1-Sin θ} + \frac{1}{1+Sin θ}$

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• LCM of (1-Sin θ) and (1 + Sin θ)

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$\sf\implies\:\frac{(1 + Sin θ) + (1 - Sin θ) }{(1-Sin θ)(1+Sin θ)}$

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• Opening the brackets

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$\sf\implies\:\frac{1 + Sin θ + 1 - Sin θ}{(1-Sin θ)(1+Sin θ)}$

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• Cancelling the same terms with opposite signs

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$\sf\implies\:\frac{1 +\cancel{ Sin θ} + 1 -\cancel{ Sin θ}}{(1-Sin θ)(1+Sin θ)}$

$\sf\implies\:\frac{1 + 1 }{(1-Sin θ)(1+Sin θ)}$

$\sf\implies\:\frac{2 }{(1-Sin θ)(1+Sin θ)}$

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• Using algebraic identity in the denominator

Identity : $\small{\underline{\boxed{\mathrm\pink{(a+b)(a-b)=a^{2} - b^{2}}}}}$

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$\sf\implies\:\frac{2 }{(1)^{2}- (Sin θ)^{2}}$

$\sf\implies\:\frac{2 }{1- Sin^{2}θ}$

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We know :

• $\tt\green{\:1-Sin^{2} θ= Cos^{2} θ}$

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$\sf\implies\:\frac{2 }{Cos^{2}θ}$

$\sf\implies\:2 \times \frac{1 }{Cos^{2}θ}$

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We know :

• $\tt\green{\:\frac{1}{Cos^{2} θ}= Sec^{2}θ}$

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$\sf\implies\:2 \times Sec^{2} θ$

$\sf\implies\:2 Sec^{2} θ$

$\sf\implies\:RHS$

$\large{\mathfrak\red{hence\:proved!!}}$

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!! Hope it helps !!