Question

tan^2 theta + cot^2 theta +2 = sec ^2 theta + cosec^2 theta

Answer 1

$\sf otan28+cotzo+2-secz0+coseco20\\\\\\otan \left(28^{\circ \:}\right)+cot \left(z\right)o+2-sec \left(z\right)\cdot \:0+csc \left(o\right)\cdot \:20\\\\\\{Aplicar\:a\:regra}\to \:0\cdot \:a=0\\\\\\=tan \left(28^{\circ \:}\right)o+ocot \left(z\right)+2-0+20csc \left(o\right)\\\\\\{Somar/subtrair:}\to\:2-0=2\\\\\\\to \boxed{\sf =tan \left(28^{\circ \:}\right)o+ocot \left(z\right)+20csc \left(o\right)+2}$

Answer 2

Answer:

=>tan^2 theta + cot^2 theta +2 = sec ^2 theta + cosec^2 theta

=>tan^2 theta + cot^2 theta +2 = 1 + tan^2 theta + 1 + cot^2 theta

=>tan^2 theta + cot^2 theta +2 = tan^2 theta + cot^2 theta +2

=>RHS = LHS

Hence Proved...

Step-by-step explanation:

Because, sec ^2 theta = 1 + tan^2 theta and cosec^2 theta = 1 + cot^2 theta.