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Sin thita ko sec thita me yakat karesin theta cos theta ​

Answer»

{sin}^{2} x + {cos}^{2} x = 1 \\ {sin}^{2} x = 1 - {cos}^{2} x \\ {sin}^{2} x = 1 - \frac{1}{ {sec}^{2} x} \: \: \: \: \: \: ... \: (cosx = \frac{1}{secx} ) \\ {sin}^{2} x = \frac{ {sec}^{2} x - 1}{ {sec}^{2}x } \\ \\ taking \: square \: roots \: on \: both \: sides \\ \\ sinx = \sqrt{ \frac{ {sec}^{2}x - 1 }{ {sec}^{2} x} } \\ sinx = \frac{ \sqrt{ {sec}^{2}x - 1 } }{secx}


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