Question

1/sec theta-tan theta-1/cos theta= 1/cos theta- 1/sec theta+tan theta

Answer 1

Proof:

Taking LHS,

$\frac{1}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta}=\frac{1}{\cos \theta}-\frac{1}{\sec \theta+\tan \theta}$

substituting$\sec ^{2} \theta-\tan ^{2} \theta=1$,

$\frac{1}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta}$

$\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta}$

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

$\sec \theta+\tan \theta-\frac{1}{\cos \theta}$

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

$\frac{\sec \theta \cos \theta+\tan \theta \cos \theta-1}{\cos \theta}$

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

$\frac{\sin \theta}{\cos \theta}=\tan \theta=L H S$

taking RHS,

$\frac{1}{\cos \theta}-\frac{1}{\sec \theta+\tan \theta}$

substituting $\sec ^{2} \theta-\tan ^{2} \theta=1,$

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

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

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

$\frac{1}{\cos \theta}-\sec \theta+\tan \theta$

$\frac{1-\cos \theta \sec \theta+\tan \theta \cos \theta}{\cos \theta}$

$\frac{1-\cos \theta \frac{1}{\cos \theta}+\tan \theta \cos \theta}{\cos \theta}$

$1-1+\tan \theta=\tan \theta=R H S$

$L H S=R H S$

Hence Proved.

Answer 2

Answer:

$\frac{1}{sec\theta-tan\theta}-\frac{1}{cos\theta}\\=\frac{1}{cos\theta}-\frac{1}{sec\theta+tan\theta}$

Step-by-step explanation:

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

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

Therefore,

$\frac{1}{sec\theta-tan\theta}-\frac{1}{cos\theta}\\=\frac{1}{cos\theta}-\frac{1}{sec\theta+tan\theta}$

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