Question

(i) Prove that $\frac{cot\theta +cosec\theta -1}{cot\theta -cosec\theta +1}=cosec\theta +cot\theta =\frac{1+cos\theta }{sin\theta }$

(ii) Prove that $\frac{1}{secA-tanA}-\frac{1}{cosA}=\frac{1}{cosA}-\frac{1}{secA+tanA}$.

Answer 1

$\left(i\right)\frac{\cot \theta +\csc \theta -1}{\cot \theta -\csc \theta +1}$

$=\frac{\cot \theta +\csc \theta -({\csc }^2\theta -{\cot }^2\theta )}{\cot \theta -\csc \theta +1}$

$=\frac{\cot \theta +\csc \theta -(\csc \theta -\cot \theta )(\csc \theta +\cot \theta )}{\cot \theta -\csc \theta +1}$

$=\frac{(\cot \theta +\csc \theta )(1-\csc \theta +\cot \theta )}{\cot \theta -\csc \theta +1}$

$=\cot \theta +\csc \theta$ .........Result$1$

$=\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta }$

$=\frac{1+\cos \theta }{\sin \theta }$ .........Result$2$

Hence from result$1$ and result$2$ we have

$\frac{\cot \theta +\csc \theta -1}{\cot \theta -\csc \theta +1}=\cot \theta +\csc \theta =\frac{1+\cos \theta }{\sin \theta }$

Hence proved.

$\left(ii\right)$L.H.S$=\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}$

Multiplying by $\sec A+\tan A$ in the numerator and denominator of first term,we get

$=\frac{\sec A+\tan A}{(\sec A-\tan A)(\sec A+\tan A)}-\sec A$

$=\frac{\sec A+\tan A}{({\sec }^{2}A-{\tan }^{2}A)}-\sec A$

$=\sec A+\tan A-\sec A$ since ${\sec }^{2}A-{\tan }^{2}A=1$

$=\tan A$

Adding and subtracting $\sec A$, we get

$=\sec A+\tan A-\sec A$

$=\frac{1}{\cos A}-(\sec A-\tan A)$

Multiplying by $\sec A+\tan A$ in the numerator and denominator of second term,we get

$=\frac{1}{\cos A}-\frac{(\sec A-\tan A)(\sec A+\tan A)}{(\sec A+\tan A)}$

$=\frac{1}{\cos A}-\frac{{\sec }^{2}A-{\tan }^{2}A}{(\sec A+\tan A)}$

$=\frac{1}{\cos A}-\frac{1}{(\sec A+\tan A)}$ since ${\sec }^{2}A-{\tan }^{2}A=1$

$=$R.H.S

$\therefore \frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{(\sec A+\tan A)}$

Hence proved.