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SSC CHSL 15 Nov 2015 Shift 2 Paper

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Question : 116 of 200

Marks: +1, -0

If cos θ =

\frac{p}{\sqrt{p^2+q^2}}

, then the value of tan is:

$\frac{q}{p}$
$\frac{p}{p^2+q^2}$
$\frac{q}{\sqrt{p^2+q^2}}$
$\frac{p}{\sqrt{p^2-q^2}}$

Solution:

Expression :

\cos \theta = \frac{p}{\sqrt{p^2+q^2}}

-----(i)

⇒

\frac{1}{\cos \theta} = \frac{\sqrt{p^2+q^2}}{p}

⇒

\sec \theta = \frac{\sqrt{p^2+q^2}}{p}

Squaring both sides, we get :

⇒

\sec^2 \theta = \left( \frac{\sqrt{p^2+q^2}}{p} \right)^2

⇒

\sec^2 \theta = \frac{p^2+q^2}{p^2}

Subtracting '1' from both sides,

⇒

\sec^2 \theta - 1 = \frac{p^2+q^2}{p^2} - 1

⇒

\tan^2 \theta = \frac{(p^2+q^2)-p^2}{p^2}

⇒

\tan^2 \theta = \frac{q^2}{p^2}

Taking square root on both sides, we get :

⇒

\sqrt{\tan^2 \theta} = \sqrt{\frac{q^2}{p^2}}

⇒

\tan \theta = \frac{q}{p}

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