Question

8) if secß + banß =p Then express The value
of sin in terms of op
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Answer 1

Correct Question:-

If sec θ + tan θ = p then express the value of sin θ in terms of p.

Answer:-

Given:

sec θ + tan θ = p -- equation (1).

We know that;

sec² θ - tan² θ = 1

using a² - b² = (a + b)(a - b) we get;

⟹ (sec θ + tan θ)(sec θ - tan θ) = 1

⟹ p(sec θ - tan θ) = 1. [∵ From equation (1) ]

⟹ sec θ - tan θ = 1/p -- equation (2).

Adding equations (1) & (2) we get;

⟹ sec θ + tan θ + sec θ - tan θ = p + 1/p

⟹ 2sec θ = (p² + 1)/p

⟹ sec θ = (p² + 1)/2p

Now;

From equation (1);

sec θ + tan θ = p

using tan θ = sin θ *(1/cos θ) in LHS we get;

⟹ sec θ (1 + sin θ) = p

[∵ 1/cos θ = sec θ]

Substitute the value of sec θ here.

⟹ (1 + sin θ) (p² + 1) / 2p = p

⟹ (1 + sin θ)(p² + 1) = 2p²

⟹ 1 + sin θ = 2p² / p² + 1

⟹ sin θ = (2p² / p² + 1) - 1

⟹ sin θ = (2p² - p² - 1) / p² + 1

⟹ sin θ = (p² - 1)/(p² + 1)

∴ The value of sin θ is p² - 1/p² + 1.