Without using trigonometric tables, evaluate each of the following:`(cos^2 20^0+cos^2 70^0)/(sec^2 50^0-cot^2 40^0)+2cos e c^2 58^0-2cot58^0tan32^0-4tan13^0tan37^0tan45^0tan53^0tan77^0`

Without using trigonometric tables, evaluate each of the following:`(c

1.	Without using trigonometric tables, evaluate each of the following:`(cos^2 20^0+cos^2 70^0)/(sec^2 50^0-cot^2 40^0)+2cos e c^2 58^0-2cot58^0tan32^0-4tan13^0tan37^0tan45^0tan53^0tan77^0`
Answer» Given expression `(cos ^2 20+cos^2 70^@)/(sec^2 50^@-cot ^2 40^@)+2cosec ^2 58^@-2cot58^@tan 32^@-4tan 13^@tan 37^@tan 45^@tan 53^@tan 77^@`. ` =(cos^2 20^@+[cos (90^@-20^@)]^2)/(sec^2 50^@-[cot(90^@-50^@)]^2)` `+2[cosec ^2 58^@-cot 58^@tan(90^@-58^@)]` ` -4 (tan 13^@tan 77^@)(tan 37^@tan 53^@)tan 45^@` `=(cos^2 20^@+sin^2 20^@)/(sec^2 50^@-tan^2 50^@)+2[cosec^2 58^@-cot^2 58^@]` ` -4tan 13^@tan(90^@-13^@).tan 37^@tan (90^@-37^@).tan 45^@ [because cos (90^@-theta)=sin theta, cot (90^@-theta) = tan theta and tan (90^@-theta )=cot theta ]` ` =(1)/(1) +(2xx 1) -4(tan 13^@cot 13^@).(tan 37^@ cot 37^@).1` `[because cos^2 theta +sin^2 theta -tan^2 theta=1, cosec^2 theta -cot^2 theta =1]` `=(3-4xx1xx1xx1)=(3-4)=-1`.