Trigonometric Identities: Diagram-Based Questions (5)

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Tags Trigonometric Identities

Question

Prove that sin⁡θ1−cos⁡θ+1−cos⁡θsin⁡θ=2csc⁡θ\tfrac{\sin\theta}{1 - \cos\theta} + \tfrac{1 - \cos\theta}{\sin\theta} = 2\csc\theta.

Solution — Step by Step

LHS=sin⁡θ1−cos⁡θ+1−cos⁡θsin⁡θ=sin⁡2θ+(1−cos⁡θ)2(1−cos⁡θ)sin⁡θ\text{LHS} = \tfrac{\sin\theta}{1 - \cos\theta} + \tfrac{1 - \cos\theta}{\sin\theta} = \tfrac{\sin^2\theta + (1 - \cos\theta)^2}{(1 - \cos\theta)\sin\theta} sin⁡2θ+(1−cos⁡θ)2=sin⁡2θ+1−2cos⁡θ+cos⁡2θ\sin^2\theta + (1 - \cos\theta)^2 = \sin^2\theta + 1 - 2\cos\theta + \cos^2\theta

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

=1+1−2cos⁡θ=2(1−cos⁡θ)= 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) LHS=2(1−cos⁡θ)(1−cos⁡θ)sin⁡θ=2sin⁡θ=2csc⁡θ\text{LHS} = \tfrac{2(1 - \cos\theta)}{(1 - \cos\theta)\sin\theta} = \tfrac{2}{\sin\theta} = 2\csc\theta

LHS == RHS, proved.

Why This Works

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is the most-used identity in trigonometry — anytime you see sin⁡2\sin^2 and cos⁡2\cos^2 together, look to combine them.

The strategy of “combine over common denominator, then simplify” works on most identity-proving questions. After combining, the numerator usually factors using a Pythagorean or double-angle identity, exposing a cancellation.

Alternative Method

Multiply numerator and denominator of the first term by 1+cos⁡θ1 + \cos\theta:

sin⁡θ(1+cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)=sin⁡θ(1+cos⁡θ)sin⁡2θ=1+cos⁡θsin⁡θ\tfrac{\sin\theta(1 + \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)} = \tfrac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \tfrac{1 + \cos\theta}{\sin\theta}

So LHS =1+cos⁡θsin⁡θ+1−cos⁡θsin⁡θ=2sin⁡θ=2csc⁡θ= \tfrac{1 + \cos\theta}{\sin\theta} + \tfrac{1 - \cos\theta}{\sin\theta} = \tfrac{2}{\sin\theta} = 2\csc\theta. Faster.

Common Mistake

Trying to “cross-multiply” identities like ordinary equations. An identity isn’t a single equation to solve — it’s a statement that LHS equals RHS. The proof is to manipulate one side until it matches the other. Cross-multiplication is fine only if you’re verifying numerically, not proving.

When stuck on an identity, try multiplying by the conjugate of any 1±cos⁡θ1 \pm \cos\theta or 1±sin⁡θ1 \pm \sin\theta term — it converts that term into sin⁡2\sin^2 or cos⁡2\cos^2 via the Pythagorean identity, often unlocking the proof in one step.

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