Question

tanA/secA-1=secA+1/tanA​

Answer 1

To prove,

$\frac{</u></strong><strong><u>tanA</u></strong><strong><u>}{</u></strong><strong><u>secA</u></strong><strong><u> - 1} = \frac{</u></strong><strong><u>secA</u></strong><strong><u> - 1}{</u></strong><strong><u>tanA</u></strong><strong><u>}$

Proof,

$</u></strong><strong><u>LHS</u></strong><strong><u> = \frac{</u></strong><strong><u>tanA</u></strong><strong><u>}{</u></strong><strong><u>secA</u></strong><strong><u> - 1} \\ = \frac{</u></strong><strong><u>tanA</u></strong><strong><u>}{</u></strong><strong><u>secA</u></strong><strong><u> - 1} \times \frac{</u></strong><strong><u>secA</u></strong><strong><u> + 1}{</u></strong><strong><u>secA</u></strong><strong><u> + 1} \\ = \frac{</u></strong><strong><u>tanA</u></strong><strong><u>(</u></strong><strong><u>secA</u></strong><strong><u> + 1)}{ {sec}^{2}</u></strong><strong><u>A</u></strong><strong><u> - 1 } \\ = \frac{</u></strong><strong><u>tanA</u></strong><strong><u>(</u></strong><strong><u>secA</u></strong><strong><u> + 1)}{ {tan}^{2} </u></strong><strong><u>A</u></strong><strong><u>} \\ = \frac{</u></strong><strong><u>secA</u></strong><strong><u> + 1}{</u></strong><strong><u>tanA</u></strong><strong><u>} \\ = </u></strong><strong><u>RHS</u></strong><strong><u>$

Hence proved☺️

(Note that,

• 1+tan²A=sec²A
• sin²A+cos²A=1
• 1+cot²A=cosec²A)

Answer 2

Answer:

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