Question

If A + B = 90° , prove that

$\sqrt{ \frac{tanAtan B + tanAcotB}{sinAsecB} - \frac{ {sin}^{2}B}{ {cos}^{2}A } } = tanA$

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

Step-by-step explanation:

Important Formulas:

(i) tan(90 - θ) = cotθ

(ii) sin(90 - θ) = cosθ

(iii) cos(90 - θ) = sinθ

(iv) cot(90 - θ) = tanθ

(v) cosecθ = (1/sinθ)

(vi) secθ = (1/cosθ)

Now,

Given, A + B = 90

Then, A = 90 - B (or) B = 90 - A.

LHS:

$Given:\sqrt{\frac{tanAtanB+tanAcotB}{sinAsecB}-\frac{sin^2B}{cos^2A}}$

$=\sqrt{\frac{tanA * tan(90-A) + tanA * cot(90-A)}{sinA * sec(90-A)} - \frac{sin^2(90-A)}{cos^2A}}$

$=\sqrt{\frac{tanAcotA+tanAtanA}{sinAcosecA}-\frac{cos^2A}{cos^2A}}$

$=\sqrt{\frac{tanA*\frac{1}{tanA}+tan^2A}{sinA*\frac{1}{sinA}}-1}$

$=\sqrt{\frac{1+tan^2A}{1}-1}$

$=\sqrt{1+tan^2A-1}$

$=\sqrt{tan^2A}$

$=tanA$

RHS

Hope it helps!