(1+tanθ+secθ)( 1+cotθ−cosec θ )=   (a) 0               (b) 2         (c) 1              (d) −1

Prove that (sec^ 2θ−1)(cosec^ 2θ−1)=1

Solve tanθ/(1-cotθ) + cotθ/(1-tanθ) = ?

(1+tan theta+sec theta)(1+cot theta-cosec theta) = ? (a) -1 (b) 0 (c) 2

The value of (cotθ+cosecθ-1)/ (cotθ-cosecθ+1) is

Prove that : (cosecθ−cotθ)^2 = (1+cosθ)/(1−cosθ) ​

Prove that : (cosecθ−cotθ)^2 = (1-cosθ)/(1+cosθ) ​

(Sec^2 θ − 1) (Cosec^2 θ − 1) =

Prove: sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ) .

Prove that (secθ−tanθ)^2 = (1-sinθ)/(1+sinθ) ​

If the angles of /_ABC are in ratio 1:1:2 respectively (the largest angle being C ),then the value of (sec A)/(cosec B)-(tan A)/(cot B) is (a) 0 (b) 1/2 (c) 1 (d) sqrt(3)/2