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Integration as Antiderivative

If tan A + ...

Question

If $tan A + sin A = m$ and $tan A + sin A = n$ then $m^{2} - n^{2} = 3 \sqrt{m n}$

A

True

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B

False

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Solution

The correct option is B False
$m^{2} - n^{2} = {(tan A + sin A)}^{2} - {(tan A + sin A)}^{2}$
$= {tan}^{2} A + {sin}^{2} A + 2 tan A sin A - {tan}^{2} A - {sin}^{2} A - 2 tan A sin A$
$= 0 \neq 3 \sqrt{m n}$

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Similar questions

tan A + sin A = m

tan A - sin A = n

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show that

m^{2} - n^{2} = 4 \sqrt{m n}

tan A + sin A = m

tan A - sin A = n

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\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = - \frac{1}{2} sec A . cos e c A = - \frac{(cot A + tan A)}{2}

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