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 }} $