JEE Main 2021 (Online) 4th August Morning ShiftQuestion 1
If $\tan A$ is equal to the integral solution of the inequality $4x^{2}-24x+35<0$ and $\sin B$ is equal to the slope of the angle bisector of the second quadrant, then $40\sin(A+B)\sin(B-A)$ is $\_\_\_\_\_\_$ .
Trigonometric Ratio & Identities
Answer$4$
Solution
Concept. Find $\tan A$ and $\sin B$, then use the product identity. Formulas used. (1) $\sin(B+A)\sin(B-A)=\sin^2B-\sin^2A$, $\sin^2A=\frac{\tan^2A}{1+\tan^2A}$Step 1 - $\tan A$. $(2x-5)(2x-7)<0$ gives $2.5<x<3.5$; the integer is $3$. So $\tan A=3$ and $\sin^2A=\frac{9}{10}$. Step 2 - $\sin B$. The bisector of the second quadrant is $y=-x$, slope $-1$, so $\sin B=-1$ and $\sin^2B=1$. Step 3 - Value. By (1), $40\left(1-\frac9{10}\right)=4$. Answer: 4