Question

answer this question


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Answer 1

✌

Step-by-step explanation:

Let, tan−1 x = α and tan−1 y = β

From tan−1 x = α we get,

x = tan α

and from tan−1 y = β we get,

y = tan β

Now, tan (α + β) = (tanα+tanβ1−tanαtanβ)

tan (α + β) = x+y1−xy

⇒ α + β = tan−1 (x+y1−xy)

⇒ tan−1 x + tan−1 y = tan−1 (x+y1−xy)

Therefore, tan−1 x + tan−1 y = tan−1 (x+y1−xy), if x > 0, y > 0 and xy < 1.

Answer 2

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${tan}^{ - 1} ( \frac{x}{y} ) - {tan}^{ - 1} ( \frac{x - y}{x + y} ) \\ = {tan}^{ - 1} ( \frac{ \frac{x}{y} - \frac{x - y}{x + y} }{1 + \frac{x}{y} . \frac{x - y}{x + y} }) \\ = {tan}^{ - 1} ( \frac{x(x + y) - y(x - y)}{y(x + y) + x(x - y)} ) \\ = {tan}^{ - 1} ( \frac{ {x}^{2} + xy - xy + {y}^{2} }{xy + {y}^{2} + {x}^{2} - xy } ) \\ = {tan}^{ - 1} ( \frac{ {x}^{2} + {y}^{2} }{ {y}^{2} + {x}^{2} } ) \\ = {tan}^{ - 1} \\ = \frac{\pi}{4}$