Prove the following with the help of identities. 1 - sin θ 1 + sin θ = sec θ - tan θ

Given, LHS = 1 - sin θ 1 + sin θ Multiplying both numerator and denominator by 1 - sin θ , = 1 - sin θ 1 + sin θ × 1 - sin θ 1 - sin θ = ( 1 - sin θ ) 2 1 - sin 2 θ = ( 1 - sin θ ) 2 cos 2 θ ∵ cos 2 θ + sin 2 θ = 1 = 1 - sin θ cos θ = 1 cos θ - sin θ cos θ = sec θ - tan θ = RHS Hence, proved.

Prove the following with the help of identities. 1 - sin θ 1 + sin θ = sec θ - tan θ

Given, LHS = 1 - sin θ 1 + sin θ Multiplying both numerator and denominator by 1 - sin θ , = 1 - sin θ 1 + sin θ × 1 - sin θ 1 - sin θ = ( 1 - sin θ ) 2 1 - sin 2 θ = ( 1 - sin θ ) 2 cos 2 θ ∵ cos 2 θ + sin 2 θ = 1 = 1 - sin θ cos θ = 1 cos θ - sin θ cos θ = sec θ - tan θ = RHS Hence, proved.

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter -1 - Trigonometry>Exercise>Q 26

EASY

10th ICSE

IMPORTANT

Earn 100

Prove the following with the help of identities.

$\sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = \sec θ - \tan θ$

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