Question

25. Prove that
1 + cos A/
sin A=
sin A/
1 - cos A​

Answer 1

Answer:

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Answer 2

ANSWER:

To Prove:

• (1 + cos A)/sin A = sin A/(1 - cos A)

Proof:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{1+\cos A}{\sin A}=\dfrac{\sin A}{1-\cos A}\\\\\text{On taking LHS,}\\\\:\implies\dfrac{1+\cos A}{\sin A}\\\\\text{Now, we will multiply and divide LHS by (1 - cos A)}\\\\\text{So,}\\\\:\implies\dfrac{1+\cos A}{\sin A}\times\dfrac{1-\cos A}{1-\cos A}\\\\:\implies\dfrac{(1+\cos A)(1-\cos A)}{\sin A(1-\cos A)}\\\\\text{We know that,}\\\\:\hookrightarrow(a+b)(a-b)=a^2-b^2\\\\\text{So,}\\\\:\implies\dfrac{(1+\cos A)(1-\cos A)}{\sin A(1-\cos A)}$

$:\implies\dfrac{1^2-\cos^2A}{\sin A(1-\cos A)}\\\\:\implies\dfrac{1-\cos^2A}{\sin A(1-\cos A)}\\\\\text{We know that,}\\\\:\hookrightarrow1-\cos^2\theta=\sin^2\theta\\\\\text{So,}\\\\:\implies\dfrac{1-\cos^2A}{\sin A(1-\cos A)}\\\\:\implies\dfrac{\sin^2A}{\sin A(1-\cos A)}\\\\\text{On cancelling sin A,}\\\\\bf{:\implies\dfrac{\sin A}{1-\cos A}=RHS}$

Formulae Used:

$:\hookrightarrow1)\:(a+b)(a-b)=a^2-b^2\\\\:\hookrightarrow2)\:1-\cos^2\theta=\sin^2\theta$

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$

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