Question

tan+tan(90-thita)=sec(sec90-thita)​

Answer 1

To prove

tan∅ + tan(90 - ∅) = sec∅sec(90 - ∅)

Taking LHS

tan∅ + tan(90 - ∅)

→ tan∅ + cot∅

(Since, tan(90 - ∅) = cot∅)

→ tan∅ + 1/tan∅

(cot∅ = 1/tan∅)

→ $\sf\frac{tan^2\theta + 1}{tan\theta}$

(By taking LCM)

→ sec²∅/tan∅

(since, tan²∅ + 1 = sec²∅)

Now, tan∅ = sec∅/cosec∅, so put tan∅ as sec∅/cosec∅

→ $\sf\frac{sec^2\theta}{\frac{sec\theta}{cosec\theta}}$

→ $\sf sec^2\theta \div \frac{sec\theta}{cosec\theta}$

→ $\sf sec^2\theta \times \frac{cosec\theta}{sec\theta}$

= sec∅cosec∅

(cancelling sec∅ by sec²∅)

→ sec∅sec(90 - ∅)

(cosec∅ = sec(90 - ∅)

Hence proved.