Question

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Answer 1

See Below.

Explanation:

We want to verify the equation:

`\displaystyle (1)/(1+\sin\theta) = \sec^2\theta - \sec\theta \tan\theta`

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

`\displaystyle (1)/(1+\sin\theta)\left((1-\sin\theta)/(1-\sin\theta)\right) = \sec^2\theta - \sec\theta \tan\theta`

Simplify. The denominator uses the difference of two squares pattern:

`\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_((a+b)(a-b)=a^2-b^2)} = \sec^2\theta - \sec\theta \tan\theta`

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

`\displaystyle \displaystyle (1-\sin\theta)/(\cos^2\theta) = \sec^2\theta - \sec\theta \tan\theta`

Split into two separate fractions:

`\displaystyle (1)/(\cos^2\theta) -(\sin\theta)/(\cos^2\theta) = \sec^2\theta - \sec\theta\tan\theta`

Rewrite the two fractions:

`\displaystyle \left((1)/(\cos\theta)\right)^2-(\sin\theta)/(\cos\theta)\cdot (1)/(\cos\theta)=\sec^2\theta - \sec\theta \tan\theta`

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

`\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta`

Hence verified.