Prove that √(sec²θ + cosec²θ) = tanθ + cotθ

Solution:

We have to prove √(sec²θ + cosec²θ) = tanθ + cotθ

By using trigonometric identities,

1 + tan² A = sec² A

cot² A + 1 = cosec² A

LHS: √(sec²θ + cosec²θ)

sec²θ = 1 + tan²θ

cosec²θ = 1 + cot²θ

So, √(sec²θ +cosec²θ) = √(1 + tan²θ + 1 + cot²θ)

= √( tan²θ + 2 + cot²θ) ------------- (1)

By using algebraic identity,

(a + b)² = a² + 2ab + b² ------------- (2)

Comparing (1) and (2)

a² = tan²θ

2ab = 2 tanθ cotθ

b² = cot²θ

So, √( tan²θ + 2 + cot²θ) = √( tan²θ + 2 tanθ cotθ + cot²θ)

We know that tanθ × cotθ = 1

Now, √( tan²θ + 2 + cot²θ) = √( tanθ + cotθ)²

√( tan²θ + 2 + cot²θ) = tanθ + cotθ

= RHS

LHS = RHS

Therefore, √(sec²θ +cosec²θ) = tanθ + cotθ

✦ Try This: Prove that sinθ(1 + tanθ) + cosθ(1 + cotθ) = (secθ + cosecθ)

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 2

Prove that √(sec²θ + cosec²θ) = tanθ + cotθ

Summary:

The cotangent of an angle in a right triangle is defined as the ratio of the adjacent side (the side adjacent to the angle) to the opposite side (the side opposite to the angle). It is proved that √(sec²θ +cosec²θ) = tanθ + cotθ

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