Question

. If tan A=n tan B, sin A=m sin B, prove that cos²A= (m²-1)/(n²-1).​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given :

$\longrightarrow\tt{tan\:A=n\:tan\:B}$

$\longrightarrow\tt{sin\:A=m\:sin\:B}$

To Prove :

$\longrightarrow\tt{{cos}^{2}\:A=\dfrac{{m}^{2}-1}{{n}^{2}-1}}$

Proof :

$\leadsto\tt{m=\dfrac{sin\:A}{sin\:B}}$

$\leadsto\tt{n=\dfrac{tan\:A}{tan\:B}}$

$\leadsto\tt{n=\dfrac{\dfrac{sin\:A}{cos\:A}}{\dfrac{sin\:B}{sin\:B}}}$

$\leadsto\tt{n=\dfrac{sin\:A\times cos\:B}{cos\:A \times sin\:B}}$

$\leadsto\tt{n=m\times\dfrac{cos\:B}{cos\:A}}$

$\leadsto\tt{n\times\:cos\:A=m\times\:cos\:B}$

Squaring both the sides.

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}\times\:{cos}^{2}\:B}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}\times\:1-{sin}^{2}\:B}$

[By cos²B = 1 - sin²B]

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}\times\:1-\dfrac{{sin}^{2}\:A}{{m}^{2}}}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}\times \dfrac{{m}^{2}-{sin}^{2}A}{{m}^{2}}}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}-{sin}^{2}A}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}-(1-{cos}^{2}\:A)}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A={m}^{2}-1+{cos}^{2}\:A}$

$\leadsto\tt{{n}^{2}\times\:{cos}^{2}\:A-{cos}^{2}\:A={m}^{2}-1}$

$\leadsto\tt{{cos}^{2}\:A\times({n}^{2}-1)={m}^{2}-1}$

$\leadsto{\boxed{\red{\tt{{cos}^{2}\:A=\dfrac{{m}^{2}-1}{{n}^{2}-1}}}}}$

$\rule{200}2$

Answer 2

⠀⠀ || ✪.SOLUTION.✪ ||

$\large{\underline{\mathrm{\bf{\:GIVEN:-}}}}$

$\mapsto\mathrm{\:(\tan A) \:=\:n\tan B}$...(1)

$\mapsto\mathrm{\:(\sin A\:=m\sin B}$..(2)

$\large{\underline{\mathrm{\bf{\:PROVE:-}}}}$

$\mapsto\mathrm{\:(\cos^2 A \:= \: \frac{(m^2-1)}{(n^2-1)}}$

$\large{\underline{\mathrm{\bf{\:PROOF:-}}}}$

by (1) :-

$\hookrightarrow\mathrm{\:\tan A \:=\:n\tan B}$

$\hookrightarrow\mathrm{\:n \:=\frac{(\tan A)}{(\tan B)}}$

$\hookrightarrow\mathrm{:n\:=\:\frac{(\frac{\sin A}{\cos A})}{(\frac{\sin B}{\cos B})}}$

$\hookrightarrow\mathrm{\:n\:=\:\frac{(\sin A\times\cos B)}{(\sin B\times\cos A)}}$

by (2) :-

$\hookrightarrow\mathrm{\:n\:=\:m\times\frac{\cos B}{\cos A}}$

$\hookrightarrow\mathrm{\:n\times\cos A\:=\:m\times\cos B}$

squaring both side,

$\hookrightarrow\mathrm{\:n^2\cos^2 A\:=\:m^2\times\cos^2 B}$

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:=\:m^2\times(1-\sin^2 B)}$

by (2) :-

$\bold{\red{\boxed{\mathrm{\sin^2 B \:=\:\frac{\sin^2 A}{m^2}}}}}$

So,

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:=\:m^2\times(1-\frac{\sin^2 A}{m^2})}$

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:=\:\cancel{m^2}\times\frac{(m^2-\sin^2 A)}{\cancel{m^2}}}$

$\bold{\pink{\boxed{\mathrm{\boxed{\:\sin^2 A\:=\:(1-\cos^2 A)}}}}}$

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:=\:m^2-(1-\cos^2 A)}$

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:=\:(m^2-1+\cos^2 A)}$

$\hookrightarrow\mathrm{\:n^2\times\cos^2 A\:-\cos^2 A \:=\:(m^2-1)}$

$\hookrightarrow\mathrm{\:\cos^2 A(n^2-1)\:=\:(m^2-1)}$

$\hookrightarrow\mathrm{\green{\boxed{\boxed{\:\cos^2 A\:=\frac{(m^2-1)}{(n^2-1)}}}}}$

$\large\underline{\mathrm{\red{\:Proved}}}$