Question

Prove the following identity: 1/secA+tanA + 1/cosA = 1/cosA - 1/secA-tanA plss answer it quickly..its very urgent The one who gives the correct answer I will surely mark that person as brainliest....

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Question: To prove,

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

Proof:

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

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

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