In a triangle ABC,if angle C = 90^{circ},then | vedclass

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

Question

(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} =

Vedclass · Accepted Answer

$\sin (A-B)$

Vedclass · Answer

(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} = k$.Then,$\frac{a^2-b^2}{a^2+b^2} = \frac{k^2 \sin^2 A - k^2 \sin^2 B}{k^2 \sin^2 A + k^2 \sin^2 B} = \frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B}$.Since $B = 90^{\circ} - A$,we have $\sin B = \cos A$ and $\cos B = \sin A$.Substituting these,$\frac{\sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A} = \frac{-(\cos^2 A - \sin^2 A)}{1} = -\cos 2A$.Alternatively,using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:$\frac{\sin(A+B)\sin(A-B)}{\sin^2 A + \cos^2 A} = \sin(90^{\circ})\sin(A-B) = 1 \cdot \sin(A-B) = \sin(A-B)$.

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

Similar Questions

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

In a triangle $ABC$,with usual notations,if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$,then $\cos A : \cos B : \cos C =$

In a $\triangle ABC$,if $a+3b=3c$,then $\sin \frac{A}{2} =$

With usual notations,the perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a=1$,then $\angle A=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module