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# solve the following identities :    1) 1+cot20/1+cosec0=cosec02)1+tan20/sec0+1=sec03)sin0(1+tan0)+cos0(1+cot0)=sec0+cosec0?4)(1+tan20)cot0/cosec20=tan0please explain me in steps.

Rinkoo Gupta
7 years ago

1+cot2ß/1+cosecß=cosecß

Cosec2ß/1+cosecß=cosecß

Cosec2ß=cosecß+cosec2ß

Cosecß=0

Sinß=8

ð No solution.

Thanks & Regards

Rinkoo Gupta

Arun Kumar IIT Delhi
7 years ago
Hi
I'll solve the first question only for rest make another post and post only one question in one post.
$\\\frac{1+cot^2\theta}{1+cosec\theta} \\\frac{sin^2\theta+cos^2\theta}{(1+cosec\theta)sin^2\theta} \\see if (1+cosec\theta) would have been just cosec\theta it would be a correct expression \\So \\=>\frac{1+cot^2\theta}{cosec\theta}=cosec\theta$

$\\1+cot^2\theta=cosec^2\theta+cosec\theta \\=>cosec\theta=0 \\which isn't possible$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty