JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

2 \cos A=x+1 / x 2 \cos B=y+1 / y

then find the value of '

k

' such that

k \cos (A-B)=\frac{x}{y}+\frac{y}{x}

View answer and explanation

Correct option: A

Option A (2) is correct because by expressing x and y in terms of cosine and sine functions and substituting into the given expression, we can simplify it to

2\cos(A-B)

, thus k=2.

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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

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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)=