Question

b cos theta=a,then prove that cosec theta+cot theta=rootb+a/b-a

Answer 1

Given :-

• $\sf{b\;cos\theta=a}$

To prove :-

• $\sf{cosec\theta+cot\theta=\sqrt{\frac{b+a}{b-a}}}$

Proof :-

As given

$\longmapsto\green{\sf{b\;cos\theta=a}}\\\\\longmapsto\green{\sf{cos\theta=\frac{a}{b}}}$

squaring both sides

$\longmapsto\green{\sf{cos^2\theta=\frac{a^2}{b^2}}}$

using identity 1 - sin²A = cos²A

$\longmapsto\green{\sf{1-sin^2\theta=\frac{a^2}{b^2}}}\\\\\longmapsto\green{\sf{sin^2\theta=1-\frac{a^2}{b^2}}}\\\\\longmapsto\green{\sf{sin^2\theta=\frac{b^2-a^2}{b^2}}}\\\\\longmapsto\green{\boxed{\sf{sin\theta=\frac{\sqrt{b^2-a^2}}{b}}}}$

Now, taking LHS

$\implies\pink{\sf{LHS=cosec\theta+cot\theta}}\\\\\implies\pink{\sf{=\frac{1}{sin\theta}+\frac{cos\theta}{sin\theta}}}\\\\\implies\pink{\sf{=\frac{1+cos\theta}{sin\theta}}}$

putting values of sinθ and cosθ

$\implies\pink{\sf{\left(1+\frac{a}{b}\right)\div\frac{\sqrt{b^2-a^2}}{b}}}\\\\\implies\pink{\sf{\frac{b+a}{b}\times \frac{b}{\sqrt{b^2-a^2}}}}\\\\\implies\pink{\sf{\frac{b+a}{\sqrt{b^2-a^2}}}}\\\\\implies\pink{\sf{\frac{(\sqrt{b+a})^2}{\sqrt{(b-a)(b+a)}}}}\\\\\implies\pink{\sf{\frac{(\sqrt{b+a})(\sqrt{b+a})}{(\sqrt{b-a})(\sqrt{b+a})}}}\\\\\implies\pink{\sf{\frac{\sqrt{b+a}}{\sqrt{b-a}}}=RHS}$

   Proved .  

Other trigonometric identities and Ratios

Trigonometric identities

$\star\;\;\boxed{\sf{sin^2A+cos^2A=1}}\\\\\star\;\;\boxed{\sf{1+tan^2A=sec^2A}}\\\\\star\;\;\boxed{\sf{1+cot^2A=cosec^2A}}$

Trigonometric ratios

$\star\;\;\boxed{\sf{cosecA=\frac{1}{sinA}}}\\\\\star\;\;\boxed{\sf{secA=\frac{1}{cosA}}}\\\\\star\;\;\boxed{\sf{tanA=\frac{sinA}{cosA}}}\\\\\star\;\;\boxed{\sf{cotA=\frac{cosA}{sinA}=\frac{1}{tanA}}}$

______________________________

$\star\;\;\boxed{\sf{sin(90-A)=cosA}}\\\star\;\;\boxed{\sf{cos(90-A)=sinA}}\\\star\;\;\boxed{\sf{tan(90-A)=cotA}}\\\star\;\;\boxed{\sf{cot(90-A)=tanA}}\\\star\;\;\boxed{\sf{sec(90-A)=cosecA}}\\\star\;\;\boxed{\sf{cosec(90-A)=secA}}$

Answer 2

Answer:

Hope it helps you.....