Trigonometric Identities: Speed-Solving Techniques (4)

easy • 3 min read
Tags Trigonometric Identities

Question

Prove that sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θ\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \dfrac{1}{\sec\theta - \tan\theta}.

Solution — Step by Step

1sec⁡θ−tan⁡θ⋅sec⁡θ+tan⁡θsec⁡θ+tan⁡θ=sec⁡θ+tan⁡θsec⁡2θ−tan⁡2θ=sec⁡θ+tan⁡θ1\frac{1}{\sec\theta - \tan\theta} \cdot \frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta} = \frac{\sec\theta + \tan\theta}{\sec^2\theta - \tan^2\theta} = \frac{\sec\theta + \tan\theta}{1}

(Using sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.)

So RHS =sec⁡θ+tan⁡θ=1+sin⁡θcos⁡θ= \sec\theta + \tan\theta = \dfrac{1 + \sin\theta}{\cos\theta}.

Divide numerator and denominator by cos⁡θ\cos\theta:

tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ\frac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}

The trick: pair −1-1 with the missing sec⁡θ\sec\theta to use sec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta.

Numerator becomes (tan⁡θ+sec⁡θ−1)(tan⁡θ+1+sec⁡θ)(\tan\theta + \sec\theta - 1)(\tan\theta + 1 + \sec\theta). Group as ((tan⁡θ+sec⁡θ)−1)((tan⁡θ+sec⁡θ)+1)=(tan⁡θ+sec⁡θ)2−1((\tan\theta + \sec\theta) - 1)((\tan\theta + \sec\theta) + 1) = (\tan\theta + \sec\theta)^2 - 1.

Denominator becomes (tan⁡θ+1−sec⁡θ)(tan⁡θ+1+sec⁡θ)=(tan⁡θ+1)2−sec⁡2θ=tan⁡2θ+2tan⁡θ+1−sec⁡2θ=2tan⁡θ(\tan\theta + 1 - \sec\theta)(\tan\theta + 1 + \sec\theta) = (\tan\theta + 1)^2 - \sec^2\theta = \tan^2\theta + 2\tan\theta + 1 - \sec^2\theta = 2\tan\theta (using sec⁡2=1+tan⁡2\sec^2 = 1 + \tan^2).

Numerator: (tan⁡θ+sec⁡θ)2−1=tan⁡2θ+2sec⁡θtan⁡θ+sec⁡2θ−1=2tan⁡2θ+2sec⁡θtan⁡θ=2tan⁡θ(tan⁡θ+sec⁡θ)(\tan\theta + \sec\theta)^2 - 1 = \tan^2\theta + 2\sec\theta\tan\theta + \sec^2\theta - 1 = 2\tan^2\theta + 2\sec\theta\tan\theta = 2\tan\theta(\tan\theta + \sec\theta).

2tan⁡θ(tan⁡θ+sec⁡θ)2tan⁡θ=tan⁡θ+sec⁡θ=sin⁡θ+1cos⁡θ\frac{2\tan\theta(\tan\theta + \sec\theta)}{2\tan\theta} = \tan\theta + \sec\theta = \frac{\sin\theta + 1}{\cos\theta}

This matches our simplified RHS. Proved.

Final answer: LHS = RHS = sec⁡θ+tan⁡θ\sec\theta + \tan\theta

Why This Works

The key identity here is sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, which lets us factor expressions of the form a2−b2a^2 - b^2 where a=sec⁡θ,b=tan⁡θa = \sec\theta, b = \tan\theta. Whenever you see sec⁡−tan⁡\sec - \tan or sec⁡+tan⁡\sec + \tan in a denominator, multiplying by the conjugate is the move.

The same template works for csc⁡±cot⁡\csc \pm \cot (since csc⁡2−cot⁡2=1\csc^2 - \cot^2 = 1).

Alternative Method

Half-angle substitution: let sin⁡θ=2t1+t2\sin\theta = \dfrac{2t}{1+t^2}, cos⁡θ=1−t21+t2\cos\theta = \dfrac{1-t^2}{1+t^2} where t=tan⁡(θ/2)t = \tan(\theta/2). Both sides reduce to rational functions of tt. Painful but mechanical.

Common Mistake

Cross-multiplying at the start: (sin⁡θ−cos⁡θ+1)(sec⁡θ−tan⁡θ)=sin⁡θ+cos⁡θ−1(\sin\theta - \cos\theta + 1)(\sec\theta - \tan\theta) = \sin\theta + \cos\theta - 1. This is fine in principle but creates a mess. Better to simplify each side separately and show they match.

Also: students often expand (sec⁡θ−tan⁡θ)2(\sec\theta - \tan\theta)^2 wrong, forgetting cross terms. The Pythagorean identity is your friend — use it.

Pattern recognition: any expression with sec⁡−tan⁡\sec - \tan in the denominator should immediately trigger “multiply by sec⁡+tan⁡\sec + \tan”. This trick alone solves 30% of trig identity proofs.

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