Question

cos sqare a-sin sqare a=1-tan sqare A÷1+tan sqare a​

Answer 1

Step-by-step explanation:

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Answer 2

$</p><p>\mathtt{Question:} \\ \cos^{2} A - \sin^{2} A = \frac{1 - \tan^{2} A }{ 1 + \tan^{2} A } \\ \\ \mathtt{Solving \; L.H.S \; separately}</p><p></p><p>\\ \\ </p><p>\rightarrow \quad \cos^{2} A - \sin^{2} A </p><p>\\ \\</p><p>\rightarrow \quad \cos^{2} A - 1 + \cos^{2} A \qquad \left( \because \; \sin^{2} A + \cos^{2} A = 1 \right)</p><p>\\ \\</p><p>\rightarrow \quad 2 \cdot \frac{1}{\sec^{2} A } - 1 \qquad \left( \because \; \sec A = \frac{1}{\cos A } \right)</p><p>\\ \\</p><p>\rightarrow \quad \frac{2 - \sec^{2} A }{\sec^{2} A } \\ \\</p><p>\rightarrow \quad \frac{2 - 1 - \tan^{2} A }{1 + \tan^{2} A} \qquad \left( \because \; 1 + \tan^{2} A = \sec^{2} A \right) </p><p>\\ \\</p><p>\rightarrow \quad \frac{1 - \tan^{2} A}{1 + \tan^{2} A} </p><p>\\ \\</p><p>\bold{ = R.H.S } </p><p>\\ \\</p><p>\bold{Hence, \: Proved} </p><p>$