# Class 10 Math Sample Paper 2019 | Question 30

#### Section D | Trigonometry

This 2019 CBSE class 10 Maths 4 mark question is from Trigonometry. Concept: Expressing trigonometric ratios in terms of sin θ and cos θ and solving a quadratic equation using the quadratic formula.

Question 30: If sec θ + tan θ = p, then find the value of cosec θ.

## NCERT Solution to Class 10 Maths

### Explanatory Answer | CBSE sample paper 2019 Question 30

#### Step 1: Rewrite sec θ + tan θ = p in terms of sin θ and cos θ

sec θ + tan θ = p
$$frac{1}{\text{cos θ}}$+ $\frac{\text{sin θ}}{\text{cos θ}}$ = p $\frac{\text{1 + sin θ }}{\text{cos θ}}$ = p 1 + sin θ = p$cos θ)

Substitute cos θ = $$sqrt{1 – {sin}^{2}θ}$ 1 + sin θ = p$\sqrt{1 – {sin}^{2}θ}$ Square both sides of the equation:$1 + sin θ)2 = p2(1 - sin2θ)
1 + sin2 θ + 2 sin θ = p2 - p2 sin2 θ
(1 + p2) sin2 θ + 2 sin θ + (1 – p2) = 0

#### Step 2: Solve the quadratic equation for sin θ

Roots of a quadratic equation ax2 + bx + c = 0 are,

$$frac{−b ± \sqrt{b^2 - 4ac}}{2a}$ Compute Discriminant, D: D = b2 - 4ac In the equation,$1 + p2) sin2 θ + 2 sin θ + (1 – p2) = θ, a = (1 + p2); b = (2); c = (1 – p2)
Discriminant, D = 4 – 4(1 + p2)(1 – p2)
= 4 – 4(1 – p4) = 4p4

Substitute the value of D,
The root, sin θ = $$frac{−2 ± √4p^4}{ 2$1 + p^2$}) = $$frac{−1 ± p^2}{$1 + p^2$}) = $\frac{p^2 − 1}{p^2 + 1}$ , -1
cosec θ = $\frac{\text{1}}{\text{sin θ}}$ = $\frac{p^2 + 1}{p^2 − 1}$ , -1

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