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Mathematics · Introduction to Trigonometry · NCERT Examples

Example 12

A complete, independently verified solution for NCERT Example 12.

Mathematics · Chapter 8 · NCERT Examples

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NCERT Example

Example 12

trigonometric identity · secant tangent

Prove that (sin θ − cos θ + 1)/(sin θ + cos θ − 1) = 1/(sec θ − tan θ), using sec²θ = 1 + tan²θ.

Solution

Board-exam working:

Divide the numerator and denominator of the left-hand side by cos θ:

LHS=tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ\mathrm{LHS}=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}
=(tan⁡θ+sec⁡θ)−1(tan⁡θ−sec⁡θ)+1=\frac{(\tan\theta+\sec\theta)-1}{(\tan\theta-\sec\theta)+1}

Multiply numerator and denominator by tan θ − sec θ:

={(tan⁡θ+sec⁡θ)−1}(tan⁡θ−sec⁡θ){(tan⁡θ−sec⁡θ)+1}(tan⁡θ−sec⁡θ)=\frac{\{(\tan\theta+\sec\theta)-1\}(\tan\theta-\sec\theta)}{\{(\tan\theta-\sec\theta)+1\}(\tan\theta-\sec\theta)}
=tan⁡2θ−sec⁡2θ−tan⁡θ+sec⁡θ(tan⁡θ−sec⁡θ+1)(tan⁡θ−sec⁡θ)=\frac{\tan^2\theta-\sec^2\theta-\tan\theta+\sec\theta}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}
=−1−tan⁡θ+sec⁡θ(tan⁡θ−sec⁡θ+1)(tan⁡θ−sec⁡θ)=\frac{-1-\tan\theta+\sec\theta}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}
=−1tan⁡θ−sec⁡θ=1sec⁡θ−tan⁡θ=RHS=\frac{-1}{\tan\theta-\sec\theta}=\frac1{\sec\theta-\tan\theta}=\mathrm{RHS}

Final answer

Hence proved.