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Math  /  Trigonometry

Question12) 11+sin⁡θ+11−sin⁡θ=2sec⁡2θ\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^{2} \theta

Studdy Solution

STEP 1

1. The equation involves trigonometric identities.
2. We will use algebraic manipulation and trigonometric identities to simplify and solve the equation.

STEP 2

1. Simplify the left-hand side of the equation.
2. Use trigonometric identities to simplify the expression.
3. Verify the equation by equating both sides.

STEP 3

Start by finding a common denominator for the left-hand side of the equation:
11+sin⁡θ+11−sin⁡θ\frac{1}{1+\sin \theta} + \frac{1}{1-\sin \theta}
The common denominator is (1+sin⁡θ)(1−sin⁡θ)(1+\sin \theta)(1-\sin \theta), which simplifies to 1−sin⁡2θ1 - \sin^2 \theta.

STEP 4

Combine the fractions over the common denominator:
11+sin⁡θ+11−sin⁡θ=1−sin⁡θ+1+sin⁡θ(1+sin⁡θ)(1−sin⁡θ)\frac{1}{1+\sin \theta} + \frac{1}{1-\sin \theta} = \frac{1-\sin \theta + 1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}
Simplify the numerator:
=21−sin⁡2θ= \frac{2}{1 - \sin^2 \theta}

STEP 5

Use the Pythagorean identity 1−sin⁡2θ=cos⁡2θ1 - \sin^2 \theta = \cos^2 \theta to simplify the expression:
2cos⁡2θ\frac{2}{\cos^2 \theta}
This simplifies to:
2sec⁡2θ2 \sec^2 \theta

STEP 6

Verify that the simplified left-hand side equals the right-hand side of the original equation:
2sec⁡2θ=2sec⁡2θ2 \sec^2 \theta = 2 \sec^2 \theta
Since both sides are equal, the equation is verified.
The equation is verified as true.

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