Question

If

${sin}^{ - 1}x + {tan}^{ - 1}x = \frac{\pi}{2} \\ find \: the \: value \: of \: {2x}^{2} + 1$
Explain the complete with steps
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Answer 1

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

Given that,

$\rm \: {sin}^{ - 1}x + {tan}^{ - 1}x = \dfrac{\pi}{2}$

Now,

Let we assume that

$\rm \: {sin}^{ - 1}x = y$

$\rm \: x = siny$

$\rm \: {x}^{2} = {sin}^{2}y$

$\rm \: {x}^{2} = 1 - {cos}^{2}y$

$\rm \:{cos}^{2}y = 1 - {x}^{2}$

$\rm \:{sec}^{2}y = \dfrac{1}{1 - {x}^{2} }$

$\rm \:1 + {tan}^{2}y = \dfrac{1}{1 - {x}^{2} }$

$\rm \: {tan}^{2}y = \dfrac{1}{1 - {x}^{2} } - 1$

$\rm \: {tan}^{2}y = \dfrac{1 - 1 + {x}^{2} }{1 - {x}^{2} }$

$\rm \: {tan}^{2}y = \dfrac{{x}^{2} }{1 - {x}^{2} }$

$\rm\implies \:tany = \dfrac{x}{ \sqrt{1 - {x}^{2} } }$

$\rm\implies \:y = {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } }$

$\rm\implies \:{sin}^{ - 1}x = {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } }$

So, given equation,

$\rm \: {sin}^{ - 1}x + {tan}^{ - 1}x = \dfrac{\pi}{2}$

can be rewritten as

$\rm \: {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } } + {tan}^{ - 1}x = \dfrac{\pi}{2}$

$\rm \: {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } } = \dfrac{\pi}{2} - {tan}^{ - 1}x$

$\rm \: {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } } = {cot}^{ - 1}x$

$\rm \: {tan}^{ - 1}\dfrac{x}{ \sqrt{1 - {x}^{2} } } = {tan}^{ - 1}\dfrac{1}{x}$

$\rm \:\dfrac{x}{ \sqrt{1 - {x}^{2} } } = \dfrac{1}{x}$

$\rm \: {x}^{2} = \sqrt{1 - {x}^{2} }$

On squaring both sides, we get

$\rm \: {x}^{4} = 1 - {x}^{2}$

$\rm \: {x}^{4} + {x}^{2} - 1 = 0$

So, by using Quadratic formula, we get

$\rm \: {x}^{2} = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)( - 1)} }{2}$

$\rm \: {x}^{2} = \dfrac{ - 1 \: \pm \: \sqrt{ 1 + 4} }{2}$

$\rm \: {x}^{2} = \dfrac{ - 1 \: \pm \: \sqrt{5} }{2}$

$\rm \: {x}^{2} = \dfrac{ - 1 \: + \: \sqrt{5} }{2} \: \: or \: \: {x}^{2} = \dfrac{ - 1 \: - \: \sqrt{5} }{2} \{rejected \: as \: {x}^{2} \cancel{ < }0 \}$

$\rm\implies \:\rm \: {x}^{2} = \dfrac{ \sqrt{5} \: - \: 1 }{2}$

So, Now Consider

$\rm \: {2x}^{2} + 1$

$\rm \: = \: 2 \times \dfrac{\sqrt{5} - 1}{2} + 1$

$\rm \: = \: \sqrt{5} - 1 + 1$

$\rm \: = \: \sqrt{5}$

Hence,

$\rm\implies \: \: \boxed{\tt{ \: {2x}^{2} + 1 = \sqrt{5} \: \: }}$

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ADDITIONAL INFORMATION

$\rm \: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} \: \: \: \: \forall \: x \: \in \: [ - 1,1]$

$\rm \: {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2} \: \: \: \: \forall \: x \: \in \: R - ( - 1,1)$

$\rm \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} \: \: \: \: \forall \: x \: \in \: R$