SSC CHSL 15 Nov 2015 Shift 2 Paper

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Question : 116 of 200

Marks: +1, -0

If cos θ =

p/√{p^2+q^2}

, then the value of tan is:

$q/p$
$p/{p^2+q^2}$
$q/√{p^2+q^2}$
$p/√{p^2-q^2}$

Solution:

Expression :

cos θ= p/√{p^2+q^2}

-----(i)
⇒

1/{cosθ} =√{p^2+q^2}/p

⇒

secθ= √{p^2+q^2}/p

Squaring both sides, we get :
⇒

sec^2θ= (√{p^2+q^2}/p)^2

⇒

sec^2θ= {p^2+q^2}/p^2

Subtracting '1' from both sides,
⇒

sec^2θ-1= {p^2+q^2}/p^2 -1

⇒

tan^2θ ={(p^2 +q^2 )−p^2}/p^2

⇒

tan^2θ =q^2/p^2

Taking square root on both sides, we get :
⇒

√{tan^2θ} =√{q^2/p^2}

⇒

tan θ =q /p

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