ABCD is a rectangle where AC is a diagonal. T | vedclass

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

Question

(C) In a rectangle $ABCD$,the diagonal $AC$ divides the rectangle into two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$.Let $\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) In a rectangle $ABCD$,the diagonal $AC$ divides the rectangle into two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$.Let $\angle CAD = 	heta$. Since $AD \parallel BC$,$\angle CAD = \angle ACB = 	heta$.Also,$\angle BAC = 90^{\circ} - 	heta$.Using the identity $1 + 	an^2 	heta = \sec^2 	heta$,the expression becomes:$(	an^2 \angle CAD + 1) \sin^2 \angle BAC = \sec^2 \angle CAD \cdot \sin^2 \angle BAC$$= \frac{1}{\cos^2 \angle CAD} \cdot \sin^2 (90^{\circ} - \angle CAD)$$= \frac{1}{\cos^2 \angle CAD} \cdot \cos^2 \angle CAD = 1$.

$ABCD$ is a rectangle where $AC$ is a diagonal. The value of $(\tan^2 \angle CAD + 1) \sin^2 \angle BAC$ is

Similar Questions

If $29 \tan \theta = 31,$ then the value of $\frac{1+2 \sin \theta \cos \theta}{1-2 \sin \theta \cos \theta}$ is equal to

The value of the following expression is $3(\sin^{4} \theta + \cos^{4} \theta) + 2(\sin^{6} \theta + \cos^{6} \theta) + 12\sin^{2} \theta \cos^{2} \theta$.

If $\tan \theta + \cot \theta = 2$,then $\sin \theta =$

Given $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$,then which of the following holds good?

In a $\Delta ABC$,$\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $D$ divides $BC$ internally in the ratio $1:3$,then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module