Question
Prove.$\frac{1}{\tan A+\cot A}=\cos A \sin A$
✓
Answer
$\frac{1}{\tan A+\cot A}=\sin A \cos A$
$\text { LHS }=\frac{1}{\tan A+\operatorname{Cot} A}$
$=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{1}{\frac{\sin ^2 A+\cos ^2}{\sin A \cos A}}$
$=\frac{1}{\frac{1}{\sin A \cos A}}\left(\because \sin ^2 A +\cos ^2 A =1\right)$
$=\sin A \cos A =\text { RHS }$
$\text { LHS }=\frac{1}{\tan A+\operatorname{Cot} A}$
$=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{1}{\frac{\sin ^2 A+\cos ^2}{\sin A \cos A}}$
$=\frac{1}{\frac{1}{\sin A \cos A}}\left(\because \sin ^2 A +\cos ^2 A =1\right)$
$=\sin A \cos A =\text { RHS }$
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
A bag contains a red ball , a blue ball and a yellow ball , all the balls being of the same size . Anjali takes out a ball from the bag without looking into it . what is the probability that she takes out .
(i) Yellow ball?
(i) Yellow ball?
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
a queen of black card
a queen of black card
Prove.$\frac{\cos A}{1-\sin A}=\sec A+\tan A$
→In the following two polynomials. Find the value of ‘a’ if x + a is a factor of each of the two:
$x^4 - a^2x^2 + 3x - a.$
→$x^4 - a^2x^2 + 3x - a.$
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears: a number divisible by 4 and 5
Two dice $($each bearing numbers $1$ to $6)$ are rolled together. Find the probability that the sum of the numbers on the upper$-$most faces of two dice is: $7, 8$ or $9$
→Prove the following identity:
$
\tan ^2 A-\sin ^2 A=\tan ^2 A \cdot \sin ^2 A
$
→$
\tan ^2 A-\sin ^2 A=\tan ^2 A \cdot \sin ^2 A
$
If $2x + 1$ is a factor of $2x^2 + ax – 3$, find the value of a.
→Find the fourth proportional to $3a, 6a^2$ and $2ab^2$
→Given: $FB = FD, AE \perp FD$ and $FC \perp AD$
Prove: $\frac{ FB }{ AD }=\frac{ BC }{ ED }$
→Prove: $\frac{ FB }{ AD }=\frac{ BC }{ ED }$