For any real values of θ , \sqrt{ \frac{\sec \\theta-1}{\sec \\theta+1}}=?

Correct Answer is : cosecθ – cotθ

SSC CGL Tier II 21 Sep 2014 Paper

Question : 40 of 100

Marks: +1, -0

For any real values of θ ,

√{ {\sec \θ-1}/{\sec \θ+1}}=?

Solution:

(3)

√{ {\sec \θ-1}/{\sec \θ+1}}= √{{1/{\cos \θ}-1}/{1/{\cos \θ}+1}}

=√{{{1-\cos \θ}/{\cos \θ} }/{{1+\cos \θ}/{\cos \θ} }}

=√{{{1-\cos \θ} }/{{1+\cos \θ} }}

=√{{({1-\cos \θ})({1-\cos \θ}) }/{({1+\cos \θ}) ({1-\cos \θ})}}

(Rationalising the numerator and the denominator)

=√{ {(1-\cos \θ)^{2}}/{1-\cos ^{2} \θ}}=√{ {(1-\cos \θ)^{2}}/{\sin ^{2} \θ}}

= {1-\cos \θ}/{\sin \θ}= {1}/{\sin \θ}- {\cos \θ}/{\sin \θ}

=\cosec \θ-\cot \θ

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