Question

hi please answer both the question I will make sure a brainlist​

Answer 1

Answer:

it is related to your question hope it helps you..

Answer 2

SOLUTION:21

Given:

sin∅ + cos∅= p

sec∅ + cosec∅= q

Now,

p² -1

=) (sin∅+cos∅)² -1

=) (sin²∅ + 2sin∅cos∅+cos²∅)-1

=) [(sin²∅+cos²∅)+2sin∅cos∅]-1

=) [1+2sin∅cos∅]-1

=) 2sin∅cos∅.............(1)

So,

q(p² -1)

=) (sec∅ +cosec∅) ×(2sin∅cos∅) [from(1)]

=) sec∅ ×(2sin∅cos∅) + cosec∅ ×(2sin∅cos∅)

=)(1/cos∅)×(2sin∅cos∅)+(1/sin∅)×(2sin∅cos∅)

=) 2sin∅ + 2cos∅

=) 2(sin∅ + cos∅)

=) 2p [proved]

SOLUTION:22

$cosec \theta + cot \theta = p...........(1) \\ now \\ cosec {}^{2} \theta - {cot}^{2} \theta = 1 \\ \\ = > (cosec \theta + cot \theta)(cosec \theta - cot \theta) = 1 \\ \\ = > p(cosec \theta - cot \theta) = 1 \\ \\ = > cosec \theta - cot \theta = \frac{1}{p}..............(2)$

Adding equation (1) & (2) we get;

$= > 2cosec \theta = p + \frac{1}{p} \\ \\ = > cosec \theta = \frac{ {p}^{2} + 1 }{2p} \\ \\ = > sin \theta = \frac{1}{cosec \theta} = \frac{2p}{ {p}^{2} + 1 } \\ \\ = > cos \theta = \sqrt{1 - {sin}^{2} \theta} = \sqrt{1 - \frac{4 {p}^{2} }{( {p}^{2} + 1) {}^{2} } } \\ \\ = > \sqrt{ \frac{ {p}^{4} + 1 - 2 {p}^{2} }{( {p}^{2} + 1) {}^{2} } } \\ \\ = > \sqrt{ \frac{( {p}^{2} - 1) {}^{2} }{( { p }^{2} + 1) {}^{2} } } \\ \\ = > \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

Proved.

Hope it helps ☺️