Question

simplify [sin theta +cosec theta whole square + [cos theta +sec theta) whole square minus tan theta + cot theta square

Answer 1

$(sin\theta +cosec\theta )^2+(cos\theta + sec\theta )^2 - (tan\theta +cot\theta )^2 =5$

Step-by-step explanation:

$(sin\theta +cosec\theta )^2+(cos\theta + sec\theta )^2 - (tan\theta +cot\theta )^2$

$=sin^2\theta +cosec^2\theta + 2sin\theta cosec\theta +cos^2\theta +sec^2\theta +2cos \theta sec\theta -tan^2\theta -cot^2\theta - 2tan\theta cot\theta$

$=(sin^2\theta+cos^2\theta) +(cosec^2\theta-cot^2\theta)+(sec^2 \theta -tan^ 2\theta )+2\frac{sin\theta }{sin\theta }+2\frac{cos\theta }{cos\theta }-2\frac{tan\theta }{tan\theta }$

=1+1+1+2+2-2

[  ∵$sin^2\theta +cos^2\theta = 1 , sec^2\theta -tan^2\theta =1\ and\ cosec^2\theta - cot^2\theta = 1$]

=5