Question

can anyone solve this question
3 and 5​

Answer 1

Let's do it.

3)

(I'm gonna take theta = alpha just for the sake of easy typing through Android)

$x = a \cos( \alpha )$
$y = b \cot( \alpha )$

I'm just gonna substitute the values I have so by using LHS,

$\frac{ {a}^{2} }{ {a}^{2} \cos {}^{2} ( \alpha ) } - \frac{ {b}^{2} }{ {b}^{2} { \cot }^{2} ( \alpha )} \\ \\ = \frac{1}{ { \cos }^{2} \alpha } - \frac{1}{ { \cot }^{2} \alpha } \\ = \frac{1 }{ { \cos}^{2} \alpha } - \frac{ \sin {}^{2} \alpha }{ { \cos}^{2} \alpha } \\ = \frac{1 - { \sin}^{2} \alpha }{ { \cos }^{2} \alpha } \\ = \frac{{ \cos }^{2} \alpha}{{ \cos }^{2} \alpha} \\ = 1$

= RHS.

5)
(Gonna take A = alpha and B = beta)

$\frac{ \tan( \alpha ) }{ \tan( \beta ) } = m \\ \\ \frac{ \sin( \alpha ) \cos( \beta ) }{ \sin( \beta ) \cos( \alpha ) } = m \\ \\ \frac{ \cos( \beta ) }{ \cos( \alpha ) } \times m = n \\ \\ \frac{ \cos( \beta ) }{ \cos( \alpha ) } = \frac{n}{m} \\ \\ m \cos( \beta ) = n \cos( \alpha )$

${n}^{2} { \cos}^{2} \alpha = {m}^{2} { \cos}^{2} \beta \\ {n}^{2} { \cos}^{2} \alpha = {m}^{2} (1 - { \sin }^{2} \beta ) \\ {n}^{2} { \cos}^{2} \alpha = {m}^{2} (1 - \frac{ { \sin }^{2} \alpha }{ {m}^{2} } ) \\ {n}^{2} { \cos}^{2} \alpha = {m}^{2} - { \sin}^{2} \alpha \\ {n}^{2} { \cos}^{2} \alpha = ( {m}^{2} - 1 + { \cos }^{2} \alpha) \\ {n}^{2} { \cos }^{2} \alpha - { \cos }^{2} \alpha = {m}^{2} - 1 \\ { \cos }^{2} \alpha ( {n}^{2} - 1) = {m}^{2} - 1 \\ \\ { \cos }^{2} \alpha = \frac{ {m}^{2} - 1 }{ {n}^{2} - 1}$

Hence proved.

Answer 2

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