Question

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Answer 1

Question:-

If $\dfrac{ \cos \alpha}{\cos \beta} = \sf m$ and $\dfrac{ \cos \alpha}{\sin \beta} = \sf n$ ,

then prove that (m² + n²) cos² β = n².

Answer:-

Given:

$\dfrac{ \cos \alpha }{ \cos \beta } = \sf m \\ \\ \dfrac{ \cos \alpha }{ \sin \beta } = \sf n$

Squaring both sides we get,

$\star \: \: \: \dfrac{ \cos ^{2} \alpha }{ \cos ^{2} \beta } = \sf m ^{2} \: \: \: - - \: (1) \\\\\star \: \: \frac{ { \cos}^{2} \alpha }{ { \sin}^{2} \beta } = \sf{n}^{2} \: \: \: - - \: (2)$

Now,

We have to prove that:

(m² + n²) cos² β = n²

Putting the values from equations (1) & (2) we get,

$\implies \sf \bigg( \dfrac{ { \cos }^{2} \alpha }{ { \cos }^{2} \beta } + \dfrac{ { { \cos }^{2} \alpha }}{ { \sin}^{2} \beta } \bigg) \times { \cos }^{2} \beta = {n}^{2} \\ \\ \\\implies \sf \bigg( \dfrac{ { \cos }^{2} \alpha \: { \sin }^{2} \beta + { \cos}^{2} \alpha \: { \cos }^{2} \beta }{ \cancel{{ \cos }^{2} \beta }\: { \sin}^{2} \beta } \bigg) \times \cancel{{ \cos }^{2} \beta} = {n}^{2} \\\\ \\ \implies \sf \frac{ { \cos }^{2} \alpha ( { \sin }^{2} \beta + { \cos}^{2} \beta )}{ { \sin}^{2} \beta } = {n}^{2}$

Using the identity sin² θ + cos² θ = 1 we get,

$\: \implies \sf \frac{ { \cos}^{2} \alpha }{ { \sin }^{2} \beta } = {n}^{2} \\ \\ \implies \boxed{\sf \: {n}^{2} = {n}^{2} }$

[ From equation (2) ]

Hence, Proved.

Answer 2

Question:-

$If \dfrac{ \cos \alpha}{\cos \beta} = \sf$

m and $\dfrac{ \cos \alpha}{\sin \beta} = \sf$ n,

• then prove that (m² + n²) cos² β = n².

$\huge\tt{\boxed{\underbrace{\overbrace{\bold\color{blue}{☛AnSwEr☚}}}}}$

Given:

$\begin{gathered}\dfrac{ \cos \alpha }{ \cos \beta } = \sf m \\ \\ \dfrac{ \cos \alpha }{ \sin \beta } = \sf n\end{gathered}$

• Squaring both sides we get,

$\begin{gathered}\star \: \: \: \dfrac{ \cos ^{2} \alpha }{ \cos ^{2} \beta } = \sf m ^{2} \: \: \: - - \: (1) \\\\\star \: \: \frac{ { \cos}^{2} \alpha }{ { \sin}^{2} \beta } = \sf{n}^{2} \: \: \: - - \: (2)\end{gathered}$

Now,

• We have to prove that:

(m² + n²) cos² β = n²

• Putting the values from equations (1) & (2) we get,

$\begin{gathered}\implies \sf \bigg( \dfrac{ { \cos }^{2} \alpha }{ { \cos }^{2} \beta } + \dfrac{ { { \cos }^{2} \alpha }}{ { \sin}^{2} \beta } \bigg) \times { \cos }^{2} \beta = {n}^{2} \\ \\ \\\implies \sf \bigg( \dfrac{ { \cos }^{2} \alpha \: { \sin }^{2} \beta + { \cos}^{2} \alpha \: { \cos }^{2} \beta }{ \cancel{{ \cos }^{2} \beta }\: { \sin}^{2} \beta } \bigg) \times \cancel{{ \cos }^{2} \beta} = {n}^{2} \\\\ \\ \implies \sf \frac{ { \cos }^{2} \alpha ( { \sin }^{2} \beta + { \cos}^{2} \beta )}{ { \sin}^{2} \beta } = {n}^{2}\end{gathered}$

• Using the identity sin² θ + cos² θ = 1 we get,

$\begin{gathered}\: \implies \sf \frac{ { \cos}^{2} \alpha }{ { \sin }^{2} \beta } = {n}^{2} \\ \\ \implies \boxed{\sf \: {n}^{2} = {n}^{2} }\end{gathered}$

[ From equation (2) ]

• Hence, Proved.