Question

Of cosec a =2,evaluate 1/tan a +sin a/[1+cos a]

Answer 1

Hello!

$\star \: \bf{cosec\:A = 2} \\ \\ \sf Then, \\ \\ \star \: \bf{sin\:A = \dfrac{1}{2}} \\ \\ \star \: \bf{cos\:A} = \sqrt{} 1 - (\dfrac{1}{2})^{2} \\ \sf cos\:A = \sqrt{} \dfrac{4-1}{4} \implies \sf cos\:A = \sqrt{} \dfrac{3}{4} \\ \\ \sf Now, \\ \\ \star \: \bf{tan\:A} = \dfrac{\sf sin\:A}{\sf cos\:A} \implies \sf tan\:A = \dfrac{1}{3} \\ \\ \underline{\sf ATQ,} \\ \\ \dfrac{1}{\sf tan\:A} + \dfrac{\sf sin\:A}{\sf 1+cos\:A} \\ \\ \implies \sqrt{3} + \dfrac{1}{2+\sqrt{3}} \\ \\ \implies \dfrac{2\sqrt{3} + 4}{2 + \sqrt{3}} \times \dfrac{2-\sqrt{3}}{2-\sqrt{3}} \\ \\ \implies \dfrac{\cancel{4\sqrt{3}}-6+8-\cancel{4\sqrt{3}}}{4-3} \\ \\ \implies \bf{8-6} \implies \boxed{\red{\bf{2}}}$

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Answer 2

$\bf \large \it Hey \: User!!!$

given cosec A = 2

we know that cosec A = hypotenuse/perpendicular

consider any ∆abc

therefore AC = 2 (hypotenuse), AB = 1 (perpendicular) and BC = ?

by Pythagoras theorem :-

>> AB² + BC² = AC²
>> 1² + BC² = 2²
>> 1 + BC² = 4
>> BC² = 4 - 1
>> BC² = 3
>> BC = √3

the base of the triangle is √3

we have to find 1/tan A + sin A/(1 + cos A)

we know that tanA = perpendicular/base
>> 1/√3

sinA = perpendicular/hypotenuse = 1/2

and cos A = base/hypotenuse = √3/2

therefore 1/tan A + sin A/( 1 + cos A)

$\sf \small \: = \frac{1}{ \frac{1}{ \sqrt{3} } } + \frac{ \frac{1}{2} }{1 + \frac{ \sqrt{3} }{2} } \\ \\ \sf \small= \frac{1}{1} \times \frac{ \sqrt{3} }{1} + \frac{ \frac{1}{2} }{ \frac{2 + \sqrt{3} }{2} } \\ \\ \sf \small= \sqrt{3} + ( \frac{1}{2} \times \frac{2}{2 + \sqrt{3} } ) \\ \\ \sf \small = \sqrt{3 } + \frac{1}{2 + \sqrt{3} } \\ \\ \sf \small = \sqrt{3} + ( \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }) \\ \\ \sf \small = \sqrt{3} + ( \frac{2 - \sqrt{3} }{ {(2})^{2} - {( \sqrt{3} )}^{2} } ) \\ \\ \sf \small = \sqrt{3} + \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \sf \small = \cancel{\sqrt{3} } + 2 - \cancel{ \sqrt{3} } \\ \\ \sf \small = \boxed{2} \: \: final \: answer$


$\bf \large \it{Cheers!!!}$