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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

We are aware of the formula on Trigonometric Inverse Function that

$\displaystyle \: {tan}^{ - 1} a + {tan}^{ - 1} b = {tan}^{ - 1} \: \frac{a + b}{1 - ab}$

2.

$\sqrt{1 + \cos 2 \theta} = \sqrt{2} \cos \theta$

3.

$\sqrt{1 - \cos 2 \theta} = \sqrt{2} \sin \theta$

TO PROVE

$\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg) = \frac{\pi}{4} + \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}$

EVALUATION

$\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg)$

Let

${x}^{2} = \cos 2 \theta$

So that

$\displaystyle \: \theta = \: \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}$

Therefore

$\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg)$

$= \displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + \cos 2 \theta} \: + \sqrt{1 - \cos 2 \theta} }{\sqrt{1 + \cos 2 \theta } \: - \sqrt{1 - \cos 2 \theta} } \: \bigg)$

$= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ \sqrt{2} \cos \theta \: + \sqrt{2} \sin \theta}{ \sqrt{2} \cos \theta \: - \sqrt{2} \sin \theta} \: \bigg)$

$= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ \cos \theta \: + \sin \theta}{ \cos \theta \: - \sin \theta} \: \bigg)$

$= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ 1 + \tan \theta}{ 1 \: - \tan \theta} \: \bigg)$

$= \displaystyle \: {tan}^{ - 1} \: 1 + {tan}^{ - 1} ( \:\tan \theta)$

$= \displaystyle \: \: \frac{\pi}{4} + \theta$

$= \displaystyle \: \: \frac{\pi}{4} + \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}$

Hence proved

Answer 2

Answer:

FORMULA TO BE IMPLEMENTED

1.

We are aware of the formula on Trigonometric Inverse Function that

\displaystyle \: {tan}^{ - 1} a + {tan}^{ - 1} b = {tan}^{ - 1} \: \frac{a + b}{1 - ab}tan

−1

a+tan

−1

b=tan

−1

1−ab

a+b

2.

\sqrt{1 + \cos 2 \theta} = \sqrt{2} \cos \theta

1+cos2θ

=

2

cosθ

3.

\sqrt{1 - \cos 2 \theta} = \sqrt{2} \sin \theta

1−cos2θ

=

2

sinθ

TO PROVE

\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg) = \frac{\pi}{4} + \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}tan

−1

(

1+x

2

−

1−x

2

1+x

2

+

1−x

2

)=

4

π

+

2

1

cos

−1

x

2

EVALUATION

\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg)tan

−1

(

1+x

2

−

1−x

2

1+x

2

+

1−x

2

)

Let

{x}^{2} = \cos 2 \thetax

2

=cos2θ

So that

\displaystyle \: \theta = \: \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}θ=

2

1

cos

−1

x

2

Therefore

\displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + {x}^{2} } \: + \sqrt{1 - {x}^{2} } }{\sqrt{1 + {x}^{2} } \: - \sqrt{1 - {x}^{2} } } \: \bigg)tan

−1

(

1+x

2

−

1−x

2

1+x

2

+

1−x

2

)

= \displaystyle \: {tan}^{ - 1} \bigg( \frac{ \sqrt{1 + \cos 2 \theta} \: + \sqrt{1 - \cos 2 \theta} }{\sqrt{1 + \cos 2 \theta } \: - \sqrt{1 - \cos 2 \theta} } \: \bigg)=tan

−1

(

1+cos2θ

−

1−cos2θ

1+cos2θ

+

1−cos2θ

)

= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ \sqrt{2} \cos \theta \: + \sqrt{2} \sin \theta}{ \sqrt{2} \cos \theta \: - \sqrt{2} \sin \theta} \: \bigg)=tan

−1

(

2

cosθ−

2

sinθ

2

cosθ+

2

sinθ

)

= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ \cos \theta \: + \sin \theta}{ \cos \theta \: - \sin \theta} \: \bigg)=tan

−1

(

cosθ−sinθ

cosθ+sinθ

)

= \displaystyle \: {tan}^{ - 1} \bigg( \: \frac{ 1 + \tan \theta}{ 1 \: - \tan \theta} \: \bigg)=tan

−1

(

1−tanθ

1+tanθ

)

= \displaystyle \: {tan}^{ - 1} \: 1 + {tan}^{ - 1} ( \:\tan \theta)=tan

−1

1+tan

−1

(tanθ)

= \displaystyle \: \: \frac{\pi}{4} + \theta=

4

π

+θ

= \displaystyle \: \: \frac{\pi}{4} + \frac{1}{2} { \cos}^{ - 1} \: {x}^{2}=

4

π

+

2

1

cos

−1

x

2

Hence proved