Question

$\mathtt \purple{ \mathtt{\boxed{Find \: the \: value : \sin ( \frac { \pi } { 2 } - \cos ^ { - 1 } \frac { 3 } { 7 } ) + \cos ( \frac { 3 \pi } { 2 } - \sin ^ { - 1 } \frac { 2 } { 7 } ) + \cos ( \tan ^ { - 1 } \frac { 7 } { 6 } )}}}$
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Answer 1

Step-by-step explanation:

Shahil Kumar roll no.14

Assessment worksheet complete science

Date 30/9/2021

worksheet no.6

Answer 2

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

Before we start solving the problem, Let's recall the formula's to be used

$\boxed{\tt{ cos( {cos}^{ - 1}x) = x \: \: \: when \: - 1 \leqslant x \leqslant 1}}$

$\boxed{\tt{ sin( {sin}^{ - 1}x) = x \: \: \: when \: - 1 \leqslant x \leqslant 1}}$

$\boxed{\tt{ sin\bigg[\dfrac{\pi}{2} - x \bigg] = cosx}}$

$\boxed{\tt{cos\bigg[\dfrac{3\pi}{2} - x \bigg] = - sinx}}$

$\boxed{\tt{ {tan}^{ - 1} \frac{y}{x} \: = \: {cos}^{ - 1} \frac{x}{ \sqrt{ {x}^{2} + {y}^{2} } \: }}}$

Let's solve the problem now!!

Consider, the expression

$\rm :\longmapsto\:\sin\bigg( \dfrac { \pi } { 2 } - \cos ^ { - 1 } \dfrac { 3 } { 7 }\bigg) + \cos \bigg( \dfrac { 3 \pi } { 2 } - \sin ^ { - 1 } \dfrac { 2 } { 7 }\bigg ) + \cos\bigg ( \tan ^ { - 1 } \dfrac { 7 } { 6 }\bigg)$

$\rm = \cos\bigg(\cos^ { - 1 } \dfrac { 3 } { 7 }\bigg) - \sin \bigg(\sin ^ { - 1 } \dfrac { 2 } { 7 }\bigg ) + \cos\bigg ( \cos^ { - 1 } \dfrac {6} { \sqrt{ {7}^{2} + {6}^{2} } }\bigg)$

$\rm \: = \: \dfrac { 3 } { 7 } - \dfrac { 2 } { 7 } + \dfrac {6} { \sqrt{ 49 + 36 }}$

$\rm \: = \: \dfrac { 3 - 2} { 7 } + \dfrac {6} { \sqrt{ 85 }}$

$\rm \: = \: \dfrac {1} { 7 } + \dfrac {6} { \sqrt{ 85 }}$

$\rm \: = \: \dfrac {1} { 7 } + \dfrac {6} { \sqrt{ 85 }} \times \dfrac{ \sqrt{85} }{ \sqrt{85} }$

$\rm \: = \: \dfrac {1} { 7 } + \dfrac {6 \sqrt{85} } {85}$

$\rm \: = \: \dfrac {85 + 42 \sqrt{85} } {595}$

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More to Know

$\boxed{\sf{ {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } ]}}$

$\boxed{\sf{ {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} } ]}}$

$\boxed{\sf{ {cos}^{ - 1}x - {cos}^{ - 1}y = {sin}^{ - 1}[x y + \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} } ]}}$

$\boxed{\sf{ {cos}^{ - 1}x + {cos}^{ - 1}y = {sin}^{ - 1}[x y - \sqrt{1 - {y}^{2} } \sqrt{1 - {x}^{2} } ]}}$

$\boxed{\sf{{2tan}^{ - 1}x = {sin}^{ - 1} \frac{2x}{1-{x}^{2} } = {tan}^{ - 1} \frac{2x}{1-{x}^{2}} = {cos}^{ - 1} \frac{1 - {x}^{2} }{1 + {x}^{2} }}}$