Question

If $15{\tan ^2}\theta + 4{\sec ^2}\theta = 23$ then ${\tan ^2}\theta =$ ……
$A.{\text{ }}\dfrac{{27}}{{15}}$
$B.{\text{ 45}}$
$C.{\text{ }}\dfrac{{19}}{{11}}$
$D.{\text{ 1}}$

Answer 1

Hint: First, we should convert ${\sec ^2}\theta$ in term of ${\tan ^2}\theta$ $\left( {{{\sec }^2}\theta = 1 + {{\tan }^2}\theta } \right)$ because we want to get the value of ${\tan ^2}\theta$ then simply solve the equation and get the value of ${\tan ^2}\theta$

Complete step-by-step answer:
$15{\tan ^2}\theta + 4{\sec ^2}\theta = 23$
Now, using the formula ${\sec ^2}\theta$ = (1+ ${\tan ^2}\theta$ ), we get
$15{\tan ^2}\theta + 4\left( {1 + {{\tan }^2}\theta } \right) = 23$
On simplifying this, we have
$15{\tan ^2}\theta + 4 + 4{\tan ^2}\theta = 23$
Now, we will take ${\tan ^2}\theta$ common, we get
$\left( {15 + 4} \right){\tan ^2}\theta +4 = 23$
Subtracting 4 on both the side,
$19{\tan ^2}\theta +4-4 = 23-4$
we get,
$19{\tan ^2}\theta = 19$

After transposing we get
${\tan ^2}\theta = \dfrac{{19}}{{19}}$
${\tan ^2}\theta = 1$
Value of ${\tan ^2}\theta$ is $1$
So, The correct option is $D$ .

Note- Some basic trigonometric equations should be in our mind which are useful for solving in this type of question
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\cos e{c^2}\theta - {\cot ^2}\theta = 1$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$