Question
Prove the following identites:
$\big(1-\cos^2\theta\big)\text{cosec}^2\theta=1$
$\big(1-\cos^2\theta\big)\text{cosec}^2\theta=1$
✓
Answer
$\text{L.H.S.}=\big(1-\cos^2\theta\big)\text{cosec}^2\theta$
$=\sin^2\theta\times\text{cosec}^2\theta$ $\Big[\because\big(1-\cos^2\theta\big)=\sin^2\theta\Big]$
$=\sin^2\theta\times\frac{1}{\sin^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
$=\sin^2\theta\times\text{cosec}^2\theta$ $\Big[\because\big(1-\cos^2\theta\big)=\sin^2\theta\Big]$
$=\sin^2\theta\times\frac{1}{\sin^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
PQRS is a trapezium with $P Q \| S R$. If $M$ and $N$ are two points on the non-parallel sides PS and QR respectively, such that MN is parallel to PQ , then show that $\frac{ PM }{ MS }=\frac{ QN }{ NR }$.
→An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=1,0^{\circ}<(A+B)<90^{\circ}$ and $A>B$ then find $A$ and $B$.
→In the given figure, LM = LN = 46°. Express x in terms of a, b and c where a, b, c are lengths of LM, MN and NK respectively.
Evaluate the following:
If A = 45°, verify that:
$\sin2\text{A}=2\sin\text{A}\cos\text{A}$
→If A = 45°, verify that:
$\sin2\text{A}=2\sin\text{A}\cos\text{A}$
Find the sum of the following APs:
9, 7, 5, 3, .... to 14 terms.
9, 7, 5, 3, .... to 14 terms.
The graph of the polynomial $f(x) = ax^2 + bx + c$ is as shown in Fig. Write the value of $b^2 - 4ac$ and the number of real zeros of f(x).
→If the distance between points $(x, 0)$ and $(0,3)$ is 5 , what are the values of $x$ ?
→Solve the following quadratic equation:$x^2 + 12x + 35 = 0$
→