Trigonometry! #15

Geometry Level 3

If 0 < A < π 4 0<A<\frac{\pi}{4} , then

1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A 1 + sin ⁡ 2 A − 1 − sin ⁡ 2 A = ? \large \frac {\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin2A}-\sqrt{1-\sin 2A}}=?

This problem is part of the set Trigonometry .

− tan ⁡ A -\tan A cot ⁡ A \cot A tan ⁡ A \tan A − cot ⁡ A -\cot A

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Chew-Seong Cheong
Jan 23, 2018

x = 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A 1 + sin ⁡ 2 A − 1 − sin ⁡ 2 A = 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A 1 + sin ⁡ 2 A − 1 − sin ⁡ 2 A × 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A = 2 + 2 ( 1 + sin ⁡ 2 A ) ( 1 − sin ⁡ 2 A ) 2 sin ⁡ 2 A = 1 + 1 − sin ⁡ 2 2 A sin ⁡ 2 A = 1 + cos ⁡ 2 A sin ⁡ 2 A = 1 + 2 cos ⁡ 2 A − 1 2 sin ⁡ A cos ⁡ A = cos ⁡ 2 A sin ⁡ A cos ⁡ A = cot ⁡ A \begin{aligned} x & = \frac {\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin 2A}-\sqrt{1-\sin 2A}} \\ & = \frac {\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin 2A}-\sqrt{1-\sin 2A}} \times \frac {\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}}{\sqrt{1+\sin 2A}+\sqrt{1-\sin 2A}} \\ & = \frac {2+2\sqrt{(1+\sin 2A)(1-\sin 2A)}}{2\sin 2A} \\ & = \frac {1+\sqrt{1-\sin^2 2A}}{\sin 2A} \\ & = \frac {1+\cos 2A}{\sin 2A} \\ & = \frac {1+2\cos^2A-1}{2\sin A\cos A} \\ & = \frac {\cos^2A}{\sin A\cos A} \\ & = \boxed{\cot A} \end{aligned}

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Omkar Kulkarni
Jan 31, 2015

Rationalise and simplify to obtain

1 + 2 cos ⁡ 2 A sin ⁡ 2 A \frac{1+2\cos 2A}{\sin 2A}

Using the double angle formula,

= 2 cos ⁡ 2 A 2 sin ⁡ A cos ⁡ A = \frac {2\cos^{2} A}{2\sin A \cos A}

= cot ⁡ A = \boxed{\cot A}

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1 + s i n 2 A = s i n 2 A + c o s 2 A + 2 s i n A c o s A = ( s i n A + c o s A ) 2 1 + sin2A = sin^2A + cos^2A + 2sinAcosA = (sinA + cosA)^2

1 − s i n 2 A = s i n 2 A + c o s 2 A − 2 s i n A c o s A = ( s i n A − c o s A ) 2 1 - sin2A = sin^2A + cos^2A - 2sinAcosA = (sinA - cosA)^2

Numerator = s i n A + c o s A + s i n A − c o s A = 2 s i n A sinA + cosA + sinA -cosA = 2sinA

denominator = s i n A + c o s A − s i n A + c o s A = 2 c o s A sinA + cosA - sinA + cosA =2cosA

the answer comes as tanA Omkar kulkarni

sandeep Rathod - 6 years, 4 months ago

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Numerator = ∣ sin ⁡ A + cos ⁡ A ∣ + ∣ sin ⁡ A − cos ⁡ A ∣ =|\sin A+\cos A|+|\sin A-\cos A|

Note that in the second term, value in it is -ve thus it will come out as cos ⁡ A − sin ⁡ A \cos A-\sin A . Same with denominator.

Pranjal Jain - 6 years, 3 months ago

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