Question

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

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Answer 1

Answer:

Step-by-step explanation

topic :

• sin / theta

to find :

• tex] \dfrac{sin theta}{1-cos theta}[/tex]= cosec

solution :

• Hope it helps the attached file

Answer 2

QUESTION-:

$\bigstar \tt \frac{Sin\; \theta}{1-cos\;\theta}=cossec\;\theta+cot\;\theta$

EXPLANATION-:

First lets solve L.H.S

$\rightarrow \bf\; \frac{Sin\; \theta}{1-cos\;\theta}$

Multiplying it by its conjugate [sin θ/1 + cos θ]

$\rightarrow \bf\; \frac{Sin\; \theta}{1-cos\;\theta}\times \frac{1+cos \;\theta}{1+cos\;\theta}\\\\\rightarrow \bf\; \frac{sin\;\theta(1+cos\;\theta)}{1-cos^2\;\theta} \;\;\;\{(a-b)(a+b)=a^2-b^2\}\\\\\rightarrow \bf\; \frac{sin\;\theta(1+cos\;\theta}{sin^2\;\theta} \;\;\{1-cos^2\;\theta=sin^2\;\theta\}\\\\\rightarrow \bf\; \frac{1+cos\;\theta}{sin\;\theta} \;\;\{canceling\;sin\;\theta\;in\;numerator\;and\;denominator\}\\\\\rightarrow \frac{1}{sin\;\theta}+\frac{1}{sin\;\theta}$

$\bf\; \\\\\longrightarrow \underline{\boxed{\ddag\;\tt cossec\;\theta+cot\;\theta\;\ddag=LHS}}\\\\\bf\;\underline{\underline{HeNcE\;\;PrOvEd}}$

EXTRA INFORMATION-:

~Trigonometric identities

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$