Question

Prove that cot square (90-theta) /tan square theta -1 + cosec square theta / sec square theta-cosec square theta=1/ sin square theta-cos square theta

Answer 1

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Answer 2

We need to recall the following trigonometric formulas.

• $tan\theta=\frac{sin\theta}{cos\theta}$
• $cot\theta=\frac{cos\theta}{sin\theta}$
• $sec\theta=\frac{1}{cos\theta}$
• $cosec\theta=\frac{1}{sin\theta}$
• $sin^2\theta+cos^2\theta=1$
• $cot^2(90-\theta)=tan\theta$

Given:

Prove that: $\frac{cot^2(90-\theta)}{tan^2\theta-1}+\frac{cosec^2\theta}{sec^2\theta-cosec^2\theta}=\frac{1}{sin^2\theta-cos^2\theta}$

Let's consider,

$\frac{cot^2(90-\theta)}{tan^2\theta-1}+\frac{cosec^2\theta}{sec^2\theta-cosec^2\theta}$

$=\frac{tan^2\theta}{tan^2\theta-1}+\frac{cosec^2\theta}{sec^2\theta-cosec^2\theta}$

$=\frac{\frac{sin^2\theta}{cos^2\theta} }{\frac{sin^2\theta}{cos^2\theta} -1}+\frac{\frac{1}{sin^2\theta} }{\frac{1}{cos^2\theta} -\frac{1}{sin^2\theta} }$

$=\frac{\frac{sin^2\theta}{cos^2\theta} }{\frac{sin^2\theta-cos^2\theta}{cos^2\theta} }+\frac{\frac{1}{sin^2\theta} }{\frac{sin^2\theta-cos^2\theta}{cos^2\theta sin^2\theta} }$

$=\frac{sin^2\theta}{sin^2\theta-cos^2\theta} +\frac{1}{\frac{sin^2\theta-cos^2\theta}{cos^2\theta} }$

$=\frac{sin^2\theta}{sin^2\theta-cos^2\theta} +\frac{cos^2\theta}{sin^2\theta-cos^2\theta} }$

$=\frac{sin^2\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

$=\frac{1}{sin^2\theta-cos^2\theta}$

Hence, proved.