If sin A, sin B, cos A are in G.P.,then the r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.Multiplying by $2$,we get $2 \sin^2 B = 2 \sin A \cos A = \si

Vedclass · Accepted Answer

Real

Vedclass · Answer

(A) Given that $\sin A, \sin B, \cos A$ are in $G.P.$,we have $\sin^2 B = \sin A \cos A$.Multiplying by $2$,we get $2 \sin^2 B = 2 \sin A \cos A = \sin 2A$.Using the identity $2 \sin^2 B = 1 - \cos 2B$,we have $1 - \cos 2B = \sin 2A$,which implies $\cos 2B = 1 - \sin 2A$.Since $-1 \le \sin 2A \le 1$,it follows that $0 \le 1 - \sin 2A \le 2$,so $\cos 2B \ge 0$.For the quadratic equation $x^2 + 2x \cot B + 1 = 0$,the discriminant $D$ is given by $D = (2 \cot B)^2 - 4(1)(1) = 4 \cot^2 B - 4 = 4(\cot^2 B - 1)$.Using $\cot^2 B - 1 = \frac{\cos^2 B - \sin^2 B}{\sin^2 B} = \frac{\cos 2B}{\sin^2 B}$,we get $D = 4 \frac{\cos 2B}{\sin^2 B}$.Since $\cos 2B \ge 0$ and $\sin^2 B > 0$,we have $D \ge 0$.Therefore,the roots of the equation are always real.

If $\sin A, \sin B, \cos A$ are in $G.P.$,then the roots of the equation $x^2 + 2x \cot B + 1 = 0$ are always

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-Px^2+Qx-R=0$ and $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are the roots of the equation $x^3-5x^2+4x=0$,then the possible least value of $P+Q+R$ is

Suppose $a, b, c$ are real numbers,and each of the equations $x^2+2ax+b^2=0$ and $x^2+2bx+c^2=0$ has two distinct real roots. Then,the equation $x^2+2cx+a^2=0$ has

For which value of $k$,are the roots of the equation $2x^2 + 5x + k = 0$ rational?

Solve the equation $2x^{2}+x+1=0$.

If for some $p, q, r \in \mathbb{R}$,not all have the same sign,and one of the roots of the equation $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$ is also a root of the equation $x^{2}+2x-8=0$,then $\frac{q^{2}+r^{2}}{p^{2}}$ is equal to-

Vedclass Test Series

Exam Paper Generator

Online Exam Module