Question

sin 0÷1-cos0=
cosec 0 +cot 0​

Answer 1

To prove :

$\dfrac{sin\theta}{1-cos\theta}=cosec\theta + cot\theta$

Proof :

Taking LHS

$\longrightarrow LHS = \dfrac{sin\theta}{1-cos\theta}$

$\red{\sf{multiplying\:by\:1+cos\theta\:in\:numerator\:and\:denominator }}$

$\longrightarrow LHS = \dfrac{sin\theta}{1-cos\theta}\times \dfrac{1+cos\theta}{1+cos\theta}$

$\red{\sf{Using\:algebraic\:identity\:a^2-b^2=(a+b)(a-b)}}$

$\longrightarrow LHS = \dfrac{sin\theta(1-cos\theta)}{(1)^2-(cos\theta)^2}$

$\longrightarrow LHS = \dfrac{sin\theta(1-cos\theta)}{1-cos^2\theta}$

$\red{\sf{using\:trigonometric\:identity\:1-cos^2A=sin^2A}}$

$\longrightarrow LHS = \dfrac{sin\theta(1-cos\theta)}{sin^2\theta}$

$\longrightarrow LHS = \dfrac{1+cos\theta}{sin\theta}$

$\longrightarrow LHS = \dfrac{1}{sin\theta} + \dfrac{cos\theta}{sin\theta}$

$\longrightarrow LHS = cosec\theta + cot\theta = RHS$

$\underline{\underline{\bf{Hence\:Proved.}}}$

know some trigonometric identites

$\red{\bigstar}\boxed{\sf{sin^A+cos^2A = 1}}$

$\red{\bigstar} \boxed{\sf{1+tan^2A =sec^2A}}$

$\red{\bigstar}\boxed{\sf{1+cot^2A = cosec^2A}}$

know some trigonometric ratios

$\red{\bigstar} \sf{ tanA= \dfrac{sinA}{cosA}}$

$\red{\bigstar}\sf{cotA = \dfrac{cosA}{sinA}}$

$\red{\bigstar}\sf{cosecA = \dfrac{1}{sinA}}$

$\red{\bigstar} \sf{ secA= \dfrac{1}{cosA}}$