If $\tan \,\,\theta = \frac{1}{{\sqrt 7 }}$, then $\frac{{\left( {{\rm{cose}}{{\rm{c}}^2}\,\theta - {{\sec }^2}\theta } \right)}}{{({\rm{cose}}{{\rm{c}}^2}\,\theta + {{\sec }^2}\theta )}}$ is equal to

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\tan \,\,\theta = \frac{1}{{\sqrt 7 }}

, then

\frac{{\left( {\mathrm{cose}{\mathrm{c}^2}\,\theta - {{\sec }^2}\theta } \right)}}{{(\mathrm{cose}{\mathrm{c}^2}\,\theta + {{\sec }^2}\theta )}}

is equal to

View answer and explanation

Correct option: B

\frac{3}{4}

Option B (

\frac{3}{4}

) is correct because it is obtained by substituting the given value of

\tan \theta

and simplifying the expression using trigonometric identities.