Question

want answer urgent give proper answer otherwise it will be reported​

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

and also pls also check that determinant sum which I posted some days ago

which of the following options doesn't equal to zero?

this one

Step-by-step explanation:

$to \: find = \\ cosec \: x + cot \: x \\ \\ given : \: \\ \\ sin \: x = \frac{2pq}{p {}^{2} + q {}^{2} } \\ \\ thus \: then \\ \\ cosec \: x = \frac{p {}^{2} + q {}^{2} }{2pq} \\ \\ we \: have \\ \\ 1 + cot {}^{2} x = cosec {}^{2} x \\ \\ cot {}^{2} x = cosec {}^{2} x - 1 \\ \\ = ( \frac{p {}^{2} + q {}^{2} }{2pq} ) {}^{2} - 1 \\ \\ = \frac{p {}^{4} + q {}^{4} + 2p {}^{2} q {}^{2} }{4p {}^{2} q {}^{2} } - 1 \\ \\ = \frac{p {}^{4} + q {}^{4} + 2p {}^{2}q {}^{2} - 4p {}^{2} q {}^{2} }{4p {}^{2}q {}^{2} } \\ \\ = \frac{p {}^{4} + q {}^{4} - 2p {}^{2} q {}^{2} }{4p {}^{2} q {}^{2} } \\ \\ = \frac{(p {}^{2} - q {}^{2}) {}^{2} }{(2pq) {}^{2} } \\ \\ = ( \frac{p {}^{2} - q {}^{2} }{2pq} ) {}^{2} \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \: then \\ \\ \\ cot \: x = \frac{p {}^{2} - q {}^{2} }{2pq} \\ \\ so \: hence \\ \\ cosec \: x + cot \: x = \frac{p {}^{2} + q {}^{2} }{2pq} + \frac{p {}^{2} - q {}^{2} }{2pq} \\ \\ = \frac{p {}^{2} +q {}^{2} + p {}^{2} - q {}^{2} }{2pq} \\ \\ = \frac{2p {}^{2} }{2pq} \\ \\ cosec \: x + cot \: x = \frac{p}{q}$