Question

sinA(1+tanA)+cosA(1+cotA) = secA+cosec​

Answer 1

Solution:-

Given

$\rm : \implies{ \sin A (1 + \tan A) + \cos A(1 + \cot A)}$

To prove

$\rm \: : \implies \: \sec A + \csc A$

Using some trigonometry identities

$\rm \to \: \tan A = \dfrac{ \sin A}{ \cos A}$

$\rm \to \cot A = \dfrac{ \cos A}{ \sin A}$

Put the value on

$\rm : \implies{ \sin A (1 + \tan A) + \cos A(1 + \cot A)}$

$\rm : \implies \sin A \bigg(1 + \dfrac{ \sin A}{ \cos A} \bigg ) + \cos A \bigg(1 + \dfrac{ \cos A}{ \sin A} \bigg)$

Taking lcm

$\rm : \implies \sin A \bigg( \dfrac{ \cos A + \sin A}{ \cos A} \bigg ) + \cos A \bigg( \dfrac{ \sin A + \cos A}{ \sin A} \bigg)$

$\rm : \implies \dfrac{ \sin A \cos A + \sin {}^{2} A}{ \cos A} + \dfrac{ \sin A \cos A + \cos {}^{2} A}{ \sin A}$

$\rm : \implies \dfrac{ \sin \: A( \sin A \cos A + \sin {}^{2} A) + \cos A(\sin A \cos A + \cos {}^{2} A)}{ \sin A \cos A}$

$\rm : \implies \dfrac{ \sin {}^{2} A \cos A + \sin {}^{3} A+ \sin A \cos {}^{2} A + \cos {}^{3} A}{ \sin A \cos A}$

$\rm : \implies \dfrac{ \sin {}^{2} A \cos A + \cos {}^{3}A + \sin {}^{3} A+ \sin A \cos {}^{2} A }{ \sin A \cos A}$

$\rm : \implies \dfrac{ \cos A (\sin {}^{2} A + \cos {}^{2}A) + \sin A( \sin {}^{2} A+ \cos {}^{2} A )}{ \sin A \cos A}$

$\rm : \implies \dfrac{ \cos A + \sin A}{ \sin A \cos A }$

$\rm : \implies \dfrac{ \cos A}{\sin A \cos A} + \dfrac{ \sin A}{\sin A \cos A}$

$\rm : \implies \: \dfrac{1}{ \sin A} + \dfrac{1}{ \cos A }$

$\rm : \implies \: \sec A + \csc A$

Hence proved

Answer 2

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