Question

Prove that cos7A/secA -sin7A/cosecA=cos 8A

Answer 1

$\huge \rm {Answer:-}$

★The solution of this question is based on one single formula,i.e,

$\mapsto \bf {cos (A+B)=cosA\: cos B-SinA\:sinB}$

★well,Let's solve and see how it goes ,

$\to \bf \red {\underline{Given:-}}$

$\to \bf {\frac{cos7A}{secA}-\frac{sin7A}{cosecA}=cos 8A}$

$\to \bf \blue {\underline{To\: prove:-}}$

$\implies{L.H.S=R.H.S}$

∵R.H.S=cos8A

$\to \bf \pink {\underline{Taking\: L.H.S:-}}$

$\to \bf {\frac{cos7A}{secA}-\frac{sin7A}{cosecA}}$

$\to \bf \purple {\underline{We\: know:-}}$

$\to \bf {cosθ=\frac{1}{secθ}}$

★so, $\to \bf {secθ=\frac{1}{cosθ}}$

★And also,

$\to \bf {sinθ=\frac{1}{cosecθ}}$

★so,$\to \bf {cosecθ=\frac{1}{sinθ}}$

★Supplanting these reciprocal identities,

$\large \to \bf {\frac{cos7A}{\frac{1}{cosA}}-\frac{sin7A}{\frac{1}{sinA}}}$

$\to \bf {cos7A\:cosA-sin7A\: sinA}$

★Now,it is in the form of

$\mapsto \bf {cos (A+B)=cosA\: cos B-SinA\:sinB}$

★A=7A and B=A

$\to \bf \orange {\underline{Thus,}}$

$\to \bf {cos(7A+A)}$

$\mapsto \bf {L.H.S=cos8A}$

$\large \implies \bf {\fbox{L.H.S=R.H.S}}$

$\to \bf \green {\underline{\underline{Hence\: Proved//}}}$

Answer 2

$\underline \bold \pink{GIVEN \: :-}$

$\bold{ \rightarrow \: \frac{ \cos \: 7a }{ \sec \: a } - \frac{ \sin \: 7a}{ \cosec \: a } = \cos \: 8a}$

$\underline \bold \blue{TO \: FIND:-}$

$\bold{ \rightarrow \: LHS \: = \: RHS \: }$

$\underline \bold \orange{SOLUTION \: :-}$

$\bold\blue{Take \: LHS,}$

$\bold{ \Rightarrow \: LHS = \frac{ \cos \: 7a }{ \sec \: a } - \frac{ \sin \: 7a}{ \cosec \: a7 } }$

$\bold \green{Write,}$

$\bold{\Rightarrow\: \frac{1}{ \:\sec \: a} = \cos \: a \:, and \frac{1}{ \csc \: a} = \sin \: a }$

$\bold{\Rightarrow \:LHS = \cos \: 7A \times \cos A = \sin 7A \times \sin A}$

$\bold \red{Using \: identity,}$

$\bold{ \cos A \: \cos B \: - \sin A \sin B = \cos(A + B)}$

$\bold{ \Rightarrow \: Here, A=7A \: and \: B = A}$

$\bold{ \Rightarrow \: LHS = cos(7A + A)}$

$\bold{ \Rightarrow \: LHS = cos(8A) \: }$

$\boxed{ \underline \bold \pink{\Rightarrow \: LHS = RHS}}$

$\underline \bold \green{Hence \: proved.}$