Question

tan^2 A - sin^2 A = tan^2 A.sin^2 A

Answer 1

Hey !!1 ^_^

Here is your answer


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${tan}^{2} a - {sin}^{2} a \\ \\ \frac{ {sin}^{2}a }{ {cos}^{2}a } - {sin}^{2}a \\ \\ {sin}^{2} a( \frac{1}{ {cos}^{2}a } - 1) \\ \\ {sin}^{2} a( \frac{1 - {cos}^{2}a }{ {cos}^{2} a} ) \\ \\ {sin}^{2} a( \frac{ {sin}^{2} a}{ {cos}^{2}a }) \\ \\ {sin}^{2} a \: . {tan}^{2} a$


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I HOPE IT WILL HELP YOU

Thank you

☺️

Answer 2

L.H.S

tan²A - sin²A,

sin²A/cos²A - sin²A,

sin²A(1/cos²A - 1),

sin²A(1-cos²A)/cos²A,

sin²A.sin²A/cos²A,

sin²A.tan²A,

hence

L.H.S=R.H.S