MHT CET · Maths · Trigonometric Equations
The number of all values of \(\theta\) in the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) satisfying the equation
\((1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2 \tan ^2 \theta=0\) is
- A 1
- B 0
- C 2
- D infinitely many.
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\((1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2 \tan ^2 \theta=0\)
\(\Rightarrow\left(1-\tan ^2 \theta\right)\left(1+\tan ^2 \theta\right)+2 \tan ^2 \theta=0\)
Put \(\tan ^2 \theta=x\)
\(\Rightarrow(1-x)(1+x)+2 x=0 \)
\( \Rightarrow 1-x^2+2 x=0 \)
\( \Rightarrow x^2-1=2 x\)
Let us draw graph of \(y=x^2-1\) and \(y=2 x\)
From the graph the two curves are intersecting at 2 points.
\(\therefore\) There are 2 values of \(x\).
Only one value of \(x\) exists, for \(\theta \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
\(\therefore x=\tan ^2 \theta\)
\(\therefore\) Two values of \(\theta\) satisfies above equation
\(\Rightarrow\left(1-\tan ^2 \theta\right)\left(1+\tan ^2 \theta\right)+2 \tan ^2 \theta=0\)
Put \(\tan ^2 \theta=x\)
\(\Rightarrow(1-x)(1+x)+2 x=0 \)
\( \Rightarrow 1-x^2+2 x=0 \)
\( \Rightarrow x^2-1=2 x\)
Let us draw graph of \(y=x^2-1\) and \(y=2 x\)
From the graph the two curves are intersecting at 2 points.
\(\therefore\) There are 2 values of \(x\).
Only one value of \(x\) exists, for \(\theta \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
\(\therefore x=\tan ^2 \theta\)
\(\therefore\) Two values of \(\theta\) satisfies above equation
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