Question

elooo oooo
Question:

[text]

In the attachment above.
answer it.

please watch radhakrishn on star bharat...!
[/ text]​

Answer 1

Answer:

$\dfrac{1+\tan ^2\mathrm{A}}{1+\cot ^2\mathrm{A}}=\dfrac{\sec ^2\left(x\right)}{\csc ^2\mathrm{A}}$

Step-by-step explanation:

$\rule{300}{1.05}$

$\dfrac{1+\tan ^2\mathrm{A}}{1+\cot ^2\mathrm{A}}$

$\rule{300}{1.05}$

$\mathrm{Use\:the\:following\:identity}:\quad \sec ^2\mathrm{A}-\tan ^2\mathrm{A}=1$

$\rule{300}{1.05}$

$\mathrm{Therefore\:}1+\tan ^2\mathrm{A}=\sec ^2\mathrm{A}$

$=\dfrac{\sec ^2\mathrm{A}}{1+\cot ^2\mathrm{A}}$

$\rule{300}{1.05}$

$\mathrm{Use\:the\:following\:identity}:\quad \:-\cot ^2\mathrm{A}+\csc ^2\mathrm{A}=1$

$\mathrm{Therefore\:}1+\cot ^2\mathrm{A}=\csc ^2\mathrm{A}$

$=\dfrac{\sec ^2\mathrm{A}}{\csc ^2\mathrm{A}}$

$\rule{300}{1.05}$

$=\dfrac{\sec^2\mathrm{A}}{1}\times\dfrac{1}{\csc^2\mathrm{A}}$

$\rule{300}{1.05}$

$=\dfrac{\sin^2\mathrm{A}}{\cos^2\mathrm{A}}$

$=\tan^2\mathrm{A}$

Answer 2

# Heya Mate #

Here is your answer

_______________________________

$\frac{1 + {tan}^{2}a }{1 + {cot}^{2} a }$

By using identities

$i \: \: \: {sec}^{2} a - {tan}^{2} a \: = 1 \\ ii \: \: {cosec}^{2} a - {cot}^{2} a = 1$

Now, by putting these identities in the question

A. T. Q

$\frac{1 + {tan}^{2}a }{ 1 + {cot}^{2}a } \\ \\ = \frac{ {sec}^{2}a }{ {cosec}^{2}a }$

Now,

Put the value of sec^2A and Cosec ^2A

which is

i. Sec^2A = 1/ Cos^2A

ii. Cosec^2A= 1/Sin^2A

Now again A.T. Q.

$\frac{1}{ {cos}^{2} a} \div \frac{1}{ {sin}^{2}a } \\ \\ now \\ \\ \frac{1}{ {cos}^{2} a} \times \frac{ {sin}^{2}a }{1} \\ \\ \frac{ {sin}^{2}a }{ {cos}^{2} a} \\ \\ = {tan}^{2} a$

In the last step we use the identity :-

Sin^2A/Cos^2A = Tan^2A

So, your answer is

${tan}^{2} a$

_____________________

Hope it helped

Mark as brainliest

:-)

✌ ✌ ✌ ✌ ThorBrainly ✌ ✌ ✌ ✌