Question

3 Prove that
sin A/(1 - cos A)=(cosecA + cot A)​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

$\sf \bullet \ \ \dfrac{sinA}{1-cosA} =(cosecA+cotA)$

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

LHS:

$\to \sf \dfrac{sinA}{1-cosA}$

⇒ Divide numerator and denominator by (1+cosA)

$\to \sf \dfrac{sinA}{1-cosA} \times \dfrac{1+cosA}{1+cosA}$

$\to \sf \dfrac{sinA(1+cosA)}{(1-cosA)(1+cosA)}$

⇒ Use (a - b)(a + b) = a² - b² to simplify denominator.

$\to \sf \dfrac{sinA(1+cosA)}{(1-cos^2A)}$

⇒ We know that 1 - cos²A = sin²A.

$\to \sf \dfrac{\cancel{sinA}(1+cosA)}{(sin^\cancel{2}A)}$

$\to \sf \dfrac{(1+cosA)}{(sinA)}$

⇒ Split the numerator.  $\sf REMEMBER:\dfrac{a+b}{c} =\dfrac{a}{c} +\dfrac{b}{c} \ OR \ \dfrac{a-b}{c} =\dfrac{a}{c} -\dfrac{b}{c}$

$\to \sf \dfrac{1}{sinA}+\dfrac{cosA}{sinA}$

⇒ We know that 1/sinA = cosecA and cosA/sinA=cotA. Therefore:

$\to \sf{\green{\dfrac{1}{sinA}+\dfrac{cosA}{sinA}=cosecA + cot A}} \bigstar$

LHS=RHS.

THUS PROVED

TRIGONOMETRIC IDENTITIES:-

$\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\  \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}$

TRIGONOMETRIC RATIOS:-

$</p><p>\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

Regards.

Sujal Sirimilla

ex-brainly star.