JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

Given that

\tan A

and

\tan B

are the roots of

x^2-p x+

q=0

, then the value of

\sin ^2(A+B)

View answer and explanation

Correct option: A

\frac{p^2}{\left\{p^2+(1-q)^2\right\}}

Option A (

\frac{p^2}{\{p^2+(1-q)^2\}}

) is correct because by using the relationship between the roots and coefficients of the quadratic equation and the tangent addition formula, the expression simplifies to the given form.

More Questions from TRIGONOMETRY RATIOS

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\tan \alpha=\frac{p}{q}

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

is equal to The value of

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

equals

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=