Question

Plz answer this I require it for m boards..

Answer 1

Solution:(Instead of θ, I use A)

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Given:

b cos A = a,

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To prove:

cosec A = cot A = $\sqrt{ \frac{b+a}{b-a} }$

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Proof:

As we know,

b cos A = a

=> $cos A = \frac{a}{b}$

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so,

We know that,

$cosecA= \frac{1}{sinA}$

&

$cotA = \frac{cosA}{sinA}$
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Hence,

cosec A + cot A

=> $\frac{1}{sinA} + \frac{cosA}{sinA}$

=> $\frac{1+cosA}{sinA}$

=> $\frac{1+( \frac{a}{b} )}{sinA}$

=> $\frac{ \frac{b+a}{b} }{ \sqrt{1 - cos^2A} }$

=> $\frac{( \frac{b+a}{b} )}{ \sqrt{1 - ( \frac{a}{b} )^2} }$

=> $\frac{( \frac{b+a}{b} )}{ \sqrt{1- \frac{a^2}{b^2} } }$

=> $( \frac{b+a}{b}) ( \frac{1}{ \sqrt{ \frac{b^2 - a^2}{b^2} } } )$

=> $( \frac{b+a}{b} )( \frac{1}{ ( \frac{ \sqrt{b^2 - a^2} }{ \sqrt{b^2} } ) } )$

=> $( \frac{b+a}{b} )( \frac{ \sqrt{b^2} }{ \sqrt{b^2 - a^2} } )$

=> $( \frac{b+a}{b} )( \frac{b}{ \sqrt{(b-a)(b+a)} } )$

=> $(b+a)( \frac{1}{( \sqrt{b+a} )( \sqrt{b-a} )} )$

=> $\frac{b+a}{( \sqrt{b+a} )( \sqrt{b-a} )}$

=> $\frac{ \sqrt{(b+a)^2} }{( \sqrt{b+a} )( \sqrt{b-a} )}$

=> $\frac{ \sqrt{(b+a)(b+a)} }{ ( \sqrt{b+a} )( \sqrt{b-a})}$

=> $\frac{ \sqrt{b+a} }{ \sqrt{b-a} }$
 
          => ∴ $\sqrt{ \frac{b+a}{b-a} }$ = RHS ,.

                    ∴ Hence proved

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                                          Hope it Helps !!