Question

3
cot (90- 0)
tan? 0-1
coseco
sec? 0 - coseco
1
sin? 0 - cos?​

Answer 1

Step-by-step explanation:

= [{cot^2(90-θ)} / (tan^2 θ - 1)] + [cosec^2 θ / (sec^2 θ - cosec^2 θ)] = 1 / (sin^2 θ - cos^2 θ);

= [{tan^2 θ} / (tan^2 θ - 1)] + [cosec^2 θ / (sec^2 θ - cosec^2 θ)] = 1 / (sin^2 θ - cos^2 θ);

(cot (90-θ) = tan θ, thus, cot^2 (90-θ) = tan^2 θ)

= {sin^2 θ/cos^2 θ} / {(sin^2 θ/cos^2 θ) - 1} + (1/sin^2 θ) / {(1/cos^2 θ) - (1/sin^2 θ)} = 1 / (sin^2 θ - cos^2 θ);

= [{sin^2 θ/cos^2 θ} / {(sin^2 θ-cos^2 θ) / cos^2 θ}] + [(1/sin^2 θ) / {(sin^2 θ-cos^2 θ)/sin^2 θ.cos^2 θ}] = 1 / (sin^2 θ - cos^2 θ);

= [sin^2 θ/(sin^2 θ-cos^2 θ)] + [ 1/{(sin^2 θ-cos^2 θ)/cos^2 θ}]= 1 / (sin^2 θ - cos^2 θ);

= sin^2 θ/(sin^2 θ-cos^2 θ) + cos^2 θ/(sin^2 θ-cos^2 θ) = 1 / (sin^2 θ - cos^2 θ);

= (sin^2 θ + cos^2 θ)/(sin^2 θ-cos^2 θ) = 1 / (sin^2 θ - cos^2 θ);

= 1/(sin^2 θ-cos^2 θ) = 1 / (sin^2 θ - cos^2 θ);

(sin^2 θ + cos^2 θ = 1)

Thus, LHS = RHS

That's all.