Question

please simplify this ​

Answer 1

Answer:

answer is 1

Step-by-step explanation:                                            $tan^{2} Q$ =$sin^{2} Q$/$cos^{2} Q$                          

let theta = Q

secQ .cosec(90-Q) - sinQ.cos(90-Q)

                                   cosQ.sin(90-Q)

secQ.secQ - sinQ.sinQ                                                        secQ=cosec(90-Q)

                     cosQ.cosQ                                                         sinQ=cos(90-Q)

                                                                                                 cosQ=sin(90-Q)

$sec^{2} Q$ - $sin^{2} Q$/$cos^{2} Q$

$sec^{2} Q$-$tan^{2} Q$           by identity $sec^{2} Q$-$tan^{2} Q$ = 1

1

Answer 2

Answer:

hope it will help you......

Step-by-step explanation:

sec theta× cosec(90-theta) - {sin theta× cos(90-theta)}/ cos theta×sin(90-theta)

as we know,

cosec(90-theta), cos(90-theta) and sin(90-theta)

is sec theta, sin theta and cos theta respectively

then,

sec theta × sec theta - {sin theta × sin theta}/{cos theta × cos theta}

sec^2 theta - ( sin^2 theta )/( cos^2 theta)

1/cos^2 theta - sin^2 theta/cos^2 theta

taking LCM,

(1 - sin^2 theta)/cos^2 theta

cos^2 theta/cos^2 theta

1

hence, the answer is 1.