Question

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Answer 1

Given, A + B = 90°

$\quad \quad A = 90\degree-B \\ \quad \quad B = 90 \degree- A$

$\text{To prove : }\sqrt{ \dfrac{ \text{tanAtan B + tanAcotB}}{ \text{sinAsecB}} - \dfrac{ { \text{sin}}^{2}B}{ \text{{cos}}^{2}A } } = \text{tanA}$

Solving left hand side,

$\implies \sqrt{ \dfrac{ \text{tanAtan B + tanAcotB}}{ \text{sinAsecB}} - \dfrac{ { \text{sin}}^{2}B}{ \text{{cos}}^{2}A } }$

From the properties of trigonometry, we know : -

=> sinA = cos( 90 - A )

=> tanA = cot( 90 - A )

$\implies \sqrt{ \dfrac{ \text{cot(90 - A)tan B + cot(90 - A)cotB}}{ \text{cos(90 - A)secB}} - \dfrac{ { \text{cos}}^{2}(90 - B)}{ \text{{sin}}^{2}A } }$

From above,

90 - A = B and 90 - B = A

$\implies \sqrt{ \dfrac{ \text{cotBtan B + cotBcotB}}{ \text{cosBsecB}} - \dfrac{ { \text{cos}}^{2}A}{ \text{{cos}}^{2}A } } \\ \\ \\ \implies \sqrt{ \dfrac{ \bigg(\dfrac{1}{tanB} \times tan B \bigg)+ cot {}^{2} B}{ { \dfrac{1}{secB} \times secB}} - 1} \\ \\ \\ \implies \sqrt{1 + \cot {}^{2}B - 1 } \\ \\ \implies \sqrt{cot {}^{2}B } \\ \\\implies cotB$

= > tan( 90 - B )

= > tanA

Hence, proved.

Answer 2

Answer is in attachment

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