Question

please solve this question​

Answer 1

QuestioN :

If sinθ = a / b then prove that ( Secθ + tan θ ) = √b + a / √ b - a.

SolutioN :

We have,

• Sin θ = a / b.

★ Let's Find Cos θ.

We know,

• sin²θ + cos²θ = 1.

So,

→ ( a / b )² + cos²θ = 1.

→ cos²θ = 1 - ( a / b )².

→ cos²θ = b² - a² / b²

→ cos² θ = √b² - a² / b.

$\rule{100}2$

Tanθ = sin θ / cos θ.

→ Tanθ = a / √b² - a².

$\rule{100}2$

Secθ = 1 / cosθ.

Secθ = b / √b² - a²

Now,

A/ Q,

→ Sec θ + tan θ.

→ b / √ b² - a² + a / √b² - a².

★ Rationalize the denominator.

→ √b + a / √ b - a

Hence Proved.

Note : Full solution provide in attachment.

Answer 2

★ DIAGRAM:

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){60}}\put(60,0){\line(0,1){45}}\put(0,0){\line(4,3){60}}\put(-4,-4){C}\put(61,-4){B}\put(61,46){A}\put(57,0.1){\framebox(3,3){}}\qbezier(5,0)(5,1)(4,3)\put(6,1){ \theta }\put(62,21){a}\put(29,25){b}\put(22,-6){ \mathrm{\sqrt{b^2-a^2}} }\end{picture}$

★ GIVEN:

• sin$\theta= \dfrac{a}{b}$

★ TO PROVE:

• secθ + tanθ$= \sqrt{\dfrac{b+a}{b-a}}$

★ SOLUTION:

cosθ = $\sf\dfrac{Adjacent}{Hypotenuse}$

cosθ = $\sf\dfrac{\sqrt{b^2 - a^2}}{b}$

secθ = $\sf\dfrac{b}{\sqrt{b^2 - a^2}}$

Assuming as equation 1

tanθ = $\sf\dfrac{sin\theta}{cos\theta} = \sf\dfrac{\frac{a}{b}}{\frac{\sqrt{b^2 - a^2}}{b}}= \sf\dfrac{a}{\sqrt{b^2 - a^2}}$ Assuming as equation 2

Adding both equations

→ secθ + tanθ = $\bigg[ \sf\dfrac{a + b}{\sqrt{b^2 - a^2}}\bigg]$

Rationalising the denominator

→ secθ + tanθ = $\sf \sqrt{\dfrac{b+a}{b-a}}$

Hence proved