Question

prove the following:

(sinA-secA)² +(cosA-cosecA)²=(1-secA.cosecA)​

Answer 1

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Answer 2

$( {sina - seca})^{2} + ( {cosa - coseca})^{2} \\ \\ = ( { \frac{sinacosa - 1}{cosa} })^{2} + ( { \frac{sinacosa - 1}{sina} })^{2} \\ \\ = \frac{( {sinacosa - 1})^{2} }{ {cos}^{2}a } + \frac{( {sinacosa - 1})^{2} }{ {sin}^{2}a } \\ \\ = ( {1 - sinacosa})^{2} ( \frac{1}{ {cos}^{2}a } + \frac{1}{ {sin}^{2} a} ) \\ \\ = ( {1 - sinacosa})^{2} ( \frac{1}{ {sin}^{2}a {cos}^{2} a } \\ \\ = ( { \frac{1 - sinacosa}{sinacosa} })^{2} \\ \\ = ( {secacoseca - 1)}^{2} \\ \\ = ( {1 - secacoseca)}^{2}$