Question

If $ \tan \alpha =\dfrac{p}{q} $ , where $ \alpha =6\beta $ , $ \alpha $ being an acute angle, prove that : $ \dfrac{1}{2}\left\{ p\operatorname{cosec}2\beta -q\sec 2\beta \right\}=\sqrt{{{p}^{2}}+{{q}^{2}}} $ .

Answer 1

Hint: To prove the above equation, first you need to convert $ \operatorname{cosec}2\beta $ and $ \sec 2\beta $ into $ \sin 2\beta $ and $ \cos 2\beta $ , using the identity $ \sin \theta =\dfrac{1}{\operatorname{cosec}\theta } $ and $ \cos \theta =\dfrac{1}{\sec \theta } $ . Then, use the identity $ \sin 2\theta =2\sin \theta \cos \theta $ and multiply numerator and denominator with $ \sqrt{{{p}^{2}}+{{q}^{2}}} $ and use triangle law of trigonometry and write $ \sin \alpha =\dfrac{p}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ and $ \cos \alpha =\dfrac{q}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ .Now, use formula $ \sin \left( A-B \right)=\sin A\cos B-\sin B\cos A $ and put $ \alpha =6\beta $ .

Complete step-by-step answer:
Since, we known that $ \sin \theta =\dfrac{1}{\operatorname{cosec}\theta } $ and $ \cos \theta =\dfrac{1}{\sec \theta } $ , then we can write LHS of the above equation $ \dfrac{1}{2}\left\{ p\operatorname{cosec}2\beta -q\sec 2\beta \right\}=\sqrt{{{p}^{2}}+{{q}^{2}}} $ as:
 $ =\dfrac{1}{2}\left\{ \dfrac{p}{\sin 2\beta }-\dfrac{q}{\cos 2\beta } \right\} $
Take LCM of $ \sin 2\beta $ and $ \cos 2\beta $ , then we will get:
 $ =\dfrac{1}{2}\left\{ \dfrac{p\cos 2\beta -q\sin 2\beta }{\sin 2\beta \cos 2\beta } \right\} $
 $ =\dfrac{p\cos 2\beta -q\sin 2\beta }{2\sin 2\beta \cos 2\beta } $
We know that $ \sin 2\theta =2\sin \theta \cos \theta $ , hence we will get
 $ =\dfrac{p\cos 2\beta -q\sin 2\beta }{\sin 4\beta } $
Now, we will multiply numerator and denominator with $ \sqrt{{{p}^{2}}+{{q}^{2}}} $ , we will get:
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\left\{ \dfrac{p\cos 2\beta -q\sin 2\beta }{\sqrt{{{p}^{2}}+{{q}^{2}}}} \right\} $
Now, split $ \sqrt{{{p}^{2}}+{{q}^{2}}} $ over both the numerator term:
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\left\{ \dfrac{p\cos 2\beta }{\sqrt{{{p}^{2}}+{{q}^{2}}}}-\dfrac{q\sin 2\beta }{\sqrt{{{p}^{2}}+{{q}^{2}}}} \right\}............\left( 1 \right) $
Now, we will use triangle law of trigonometry (i.e. $ \sin \theta =\dfrac{perpendicular}{hypotenuse} $ , $ \cos \theta =\dfrac{base}{hypotenuse} $ and \[\tan \theta =\dfrac{perpendicular}{base}\])
It is given in question that $ \tan \alpha =\dfrac{p}{q} $ , hence perpendicular of the triangle is ‘p’ and its base is ‘q’, then, the hypotenuse will become $ \sqrt{{{p}^{2}}+{{q}^{2}}} $ , so we can draw the below diagram:

From the figure, it can be seen that we can write $ \sin \alpha =\dfrac{p}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ and $ \cos \alpha =\dfrac{q}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ .
Hence, the above equation (1) will become:
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\left\{ \sin \alpha \cos 2\beta -\cos \alpha \sin 2\beta \right\} $
Now, by using the formula $ \sin \left( A-B \right)=\sin A\cos B-\sin B\cos A $ , we can rewrite the above equation as:
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\sin \left( \alpha -2\beta \right) $
Now, we will put $ \alpha =6\beta $ in the above equation, then we will get:
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\sin \left( 6\beta -2\beta \right) $
 $ =\dfrac{\sqrt{{{p}^{2}}+{{q}^{2}}}}{\sin 4\beta }\sin \left( 4\beta \right) $
 $ =\sqrt{{{p}^{2}}+{{q}^{2}}} $ = RHS
Hence, LHS = RHS
This is our required proof.

Note: Students are required to check that weather $ \alpha $ is an acute angle or not, otherwise there is chance of change of sign while writing $ \sin \alpha =\dfrac{p}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ and $ \cos \alpha =\dfrac{q}{\sqrt{{{p}^{2}}+{{q}^{2}}}} $ , if $ \alpha $ become greater than $ 90{}^\circ $ .