Question

prove that (1÷sec thita-tan theta)-(1÷cos theta)=(1÷cos theta)-(1÷sec theta+tan theta)​

Answer 1

Answer:

Step-by-step explanation:

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

Step-by-step explanation:

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