Question

Please solve questions 17

Answer 1

$\underline{\underline{\Large{\mathfrak{Solution : }}}}$



$\underline{\textsf{Given,}} \\ \\ \mathsf{\implies cos( \alpha \: + \: \beta ) \: = \: 0 }$



$\textsf{We know that, } \\ \\ \mathsf{\implies cos \: 90^{\circ} \: = \: 0 }$



$\textsf{Now,} \\ \\ \mathsf{\implies cos( \alpha \: + \: \beta )\: = \: cos \: 90^{\circ} } \\ \\ \mathsf{\implies \alpha \: + \: \beta \: = \: 90^{\circ} \qquad...(1)}$



$\textsf{Now,} \\ \\ \mathsf{= sin( \alpha \: - \: \beta) } \\ \\ \mathsf{ = cos \{90 ^{\circ} \: - \: ( \alpha \: - \: \beta) \}} \\ \\ \textsf{Plug the value of ( 1 ),} \\ \\ \mathsf{= cos \{( \alpha \: + \: \beta) \: - \: ( \alpha \: - \: \beta )\}} \\ \\ \mathsf{ = cos( \cancel{\alpha }\: + \: \beta \: - \: \cancel{ \alpha} \: + \: \beta)} \\ \\ \mathsf{ = cos \: 2 \beta}$


$\textsf{Trigonometric identity used : } \\ \\ \mathsf{ \implies cos(90^{\circ} \: - \: \theta ) \: = \: sin \: \theta }$

Answer 2

$\underline{\underline{\Large{\mathfrak{Solution : }}}} \\\\\\</p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\underline{\textsf{Given,}} \\ \\ \mathsf{\implies cos( \alpha \: + \: \beta ) \: = \: 0 }\end{lgathered} </p><p>\\\\\\</p><p> </p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{We know that, } \\ \\ \mathsf{\implies cos \: 90^{\circ} \: = \: 0 }\end{lgathered} </p><p>\\\\\\</p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{Now,} \\ \\ \mathsf{\implies cos( \alpha \: + \: \beta )\: = \: cos \: 90^{\circ} } \\ \\ \mathsf{\implies \alpha \: + \: \beta \: = \: 90^{\circ} \qquad...(1)}\end{lgathered} </p><p></p><p> \\\\\\</p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{Now,} \\ \\ \mathsf{= sin( \alpha \: - \: \beta) } \\ \\ \mathsf{ = cos \{90 ^{\circ} \: - \: ( \alpha \: - \: \beta) \}} \\ \\ \textsf{Plug the value of ( 1 ),} \\ \\ \mathsf{= cos \{( \alpha \: + \: \beta) \: - \: ( \alpha \: - \: \beta )\}} \\ \\ \mathsf{ = cos( \cancel{\alpha }\: + \: \beta \: - \: \cancel{ \alpha} \: + \: \beta)} \\ \\ \mathsf{ = cos \: 2 \beta}\end{lgathered} </p><p></p><p> \\\\\\</p><p> </p><p></p><p></p><p>\begin{lgathered}\textsf{Trigonometric identity used : } \\ \\ \mathsf{ \implies cos(90^{\circ} \: - \: \theta ) \: = \: sin \: \theta }\end{lgathered}$