IIT JEE Main Maths -Unit 14 -Trigonometric identities and trigonometric functions- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Ans. (3)
Sol. \( f(x) = (7\tan^6 x – 3\tan^2 x)(\sec^2 x) \)
\( I_1 = \int_{0}^{\pi/4} (7\tan^6 x – 3\tan^2 x)(\sec^2 x) \, dx \)
Put \( \tan x = t \)
\( I_1 = \int_{0}^{1} (7t^6 – 3t^2) \, dt = \left[ t^7 – t^3 \right]_{0}^{1} =0 \)
\( I_2 = \int_{0}^{\pi/4} x (7\tan^6 x – 3\tan^2 x)(\sec^2 x) \, dx \)
\( = \int_{0}^{\pi/4} x (\tan^7 x – \tan^3 x) \, dx – \int_{0}^{\pi/4} (\tan^7 x – \tan^3 x) \, dx \)
\( = \int_{0}^{\pi/4} \tan^3 x (\tan^2 x – 1)(-1 + \tan^2 x) \, dx \)
Put \( \tan x = t \)
\( = \int_{0}^{1} (t^5 – t^3 – t^3 + t) \, dt = \left[ -\frac{t^6}{6} + \frac{t^4}{4} – \frac{t^4}{4} + \frac{t^2}{2} \right]_{0}^{1} = -\frac{1}{6} + \frac{1}{4} – \frac{1}{4} + \frac{1}{2} = \frac{1}{12} \)
\( 7I_1 + 12I_2 = 7 \times 0 + 12 \times \frac{1}{12} = 1 \)
▶ Answer/Explanation
Sol. $2\sin^2 2\theta = \cos 2\theta$
$2\sin^2 2\theta = 1 – 2\sin^2 2\theta$
$4\sin^2 2\theta = 1$
$\sin 2\theta = \frac{1}{2}$
$\sin \theta = \pm \frac{1}{2}$
$2\cos^2 2\theta = 3\sin \theta$
$2 – 2\sin^2 2\theta + 3\sin \theta – 2 = 0$
$(2\sin \theta – 1)(2\sin \theta – 2) = 0$
$\sin \theta = \frac{1}{2}$
so common equation which satisfy both equations is $\sin \theta = \frac{1}{2}$
$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ ($\theta \in [0, 2\pi]$)
Sum = $\pi$
▶️ Answer/Explanation
Ans. (1)
Sol. \(\sin 70^\circ (\cot 10^\circ \cot 70^\circ – 1)\)
\(\Rightarrow \frac{\cos 80^\circ}{\sin 10^\circ} = 1\)