Question

is sin^-1 4/5 + sin^-1 12/13 + sin^-1 33/65 = π/2​

Answer 1

Answer:

$\bold{Yes \: statement \: is \: true}$

Step-by-step explanation:

$\bold{To \: verify}\longrightarrow \\ {sin}^{ - 1} ( \frac{4}{5} ) \: + {sin}^{ - 1} ( \frac{12}{13} ) + {sin}^{ - 1} ( \frac{33}{65} ) = \dfrac{\pi}{2}$

$\bold{Concept \: used}\longrightarrow \\ 1) {sin}^{ - 1} x + {sin}^{ - 1} y = {sin}^{ - 1} (x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } \: )$

$2) {sin}^{ - 1} x = {cos}^{ - 1} \sqrt{1 - {x}^{2} }$

$3) {sin}^{ - 1} x + {cos}^{ - 1} x = \dfrac{\pi}{2}$

$\bold{Solution}\longrightarrow \\ {sin}^{ - 1} ( \dfrac{4}{5} ) + {sin}^{ - 1} ( \dfrac{12}{13} ) + {sin}^{ - 1} ( \dfrac{33}{65} )$

$= {sin}^{ - 1} ( \: \dfrac{4}{5} \sqrt{1 - {( \dfrac{12}{13} )}^{2} } + \dfrac{12}{13} \sqrt{1 - {( \dfrac{4}{5}) }^{2} } \: ) + {cos}^{ - 1} \sqrt{1 - {( \dfrac{33}{65}) }^{2} }$

$= {sin}^{ - 1} ( \: \dfrac{4}{5} \sqrt{1 - \dfrac{144}{169} } + \dfrac{12}{13} \sqrt{1 - \dfrac{16}{25} } \: ) + {cos}^{ - 1} \sqrt{1 - \dfrac{1089}{4225} }$

$= {sin}^{ - 1} ( \: \dfrac{4}{5} \: \dfrac{169 - 144}{169} \: + \dfrac{12}{13} \: \dfrac{25 - 16}{25} \: ) + {cos}^{ - 1} \sqrt{ \dfrac{4225 - 1089}{4225} }$

$= {sin}^{ - 1} ( \: \dfrac{4}{5} \: \dfrac{5}{13} + \dfrac{12}{13} \: \dfrac{3}{5} \: ) + {cos}^{ - 1} \sqrt{ \dfrac{3136}{4225} }$

$= {sin}^{ - 1} ( \dfrac{20 + 36}{65} ) + {cos}^{ - 1} ( \dfrac{56}{65} )$

$= {sin}^{ - 1} ( \dfrac{56}{65} ) + {cos}^{ - 1} ( \dfrac{56}{65} )$

$= \dfrac{\pi}{2}$

$it \: means \: given \: relation \: is \: true$