If frac{tan(A-B)}{tan A} + frac{sin^{2}C}{sin | vedclass

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

Question

(A) Given: $\frac{	an(A-B)}{	an A} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$, we have: $\frac

Vedclass · Accepted Answer

$	an A, 	an C, 	an B$ are in $G$.$P$.

Vedclass · Answer

(A) Given: $\frac{	an(A-B)}{	an A} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$, we have: $\frac{	an A - 	an B}{	an A(1 + 	an A 	an B)} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Let $	an A = x, 	an B = y, 	an C = z$. Since $\sin^{2}C = \frac{	an^{2}C}{1 + 	an^{2}C} = \frac{z^{2}}{1 + z^{2}}$ and $\sin^{2}A = \frac{x^{2}}{1 + x^{2}}$, the equation becomes: $\frac{x-y}{x(1+xy)} + \frac{z^{2}(1+x^{2})}{x^{2}(1+z^{2})} = 1$ Multiplying by $x^{2}(1+xy)(1+z^{2})$ and simplifying: $x(x-y)(1+z^{2}) + z^{2}(1+x^{2})(1+xy) = x^{2}(1+xy)(1+z^{2})$ After algebraic expansion and cancellation, we obtain: $z^{2} = xy$ $	herefore 	an^{2}C = 	an A \cdot 	an B$ Thus, $	an A, 	an C, 	an B$ are in $G$.$P$.

If $\frac{\tan(A-B)}{\tan A} + \frac{\sin^{2}C}{\sin^{2}A} = 1,$ where $A, B, C \in (0, \frac{\pi}{2})$, then:

Similar Questions

If $\tanh x = \operatorname{sech} y = \frac{3}{5}$ and $e^{x+y}$ is an integer,then $e^{x+y} =$

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

If $\sinh u = \tan \theta$,then $\cosh u$ is equal to

If $a \cos 2\theta + b \sin 2\theta = c$ has $\alpha$ and $\beta$ as its solutions,then the value of $\tan \alpha + \tan \beta$ is

If $O$ is the circumcentre of the $\Delta ABC$ and $R_1, R_2$ and $R_3$ are the radii of the circumcircles of triangles $OBC, OCA$ and $OAB$ respectively,then the value of $\frac{a}{R_1} + \frac{b}{R_2} + \frac{c}{R_3}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module