We have cot A=cos A/sin A=2 cos^2 A/2 sin A cos A=1+cos 2A/sin 2A Putting A=71^∘/2⇒ cot 71^∘/2=1+cos 15^∘/sin 15^∘ On simplification, We get cot 71^∘/2=√(6)+√(2 ...

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Standard XI

Mathematics

Trigonometric Ratios for Sum of Two Angles

√2+√3+√4+√6 i...

Question

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

c o t 7 \frac{1^{\circ}}{2}

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s i n 7 \frac{1^{\circ}}{2}

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s i n 15^{\circ}

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c o s 15^{\circ}

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Solution

The correct option is A $c o t 7 \frac{1^{\circ}}{2}$
We have $c o t A = \frac{c o s A}{s i n A} = \frac{2 c o s^{2} A}{2 s i n A c o s A} = \frac{1 + c o s 2 A}{s i n 2 A}$
Putting $A = 7 \frac{1^{\circ}}{2} \Rightarrow c o t 7 \frac{1^{\circ}}{2} = \frac{1 + c o s 15^{\circ}}{s i n 15^{\circ}}$
On simplification,
We get $c o t 7 \frac{1^{\circ}}{2} = \sqrt{6} + \sqrt{2} + \sqrt{3} + \sqrt{4} .$

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√2+√3+√4+√6 is equal to

The correct option is A cot 71∘2We have cot A=cos Asin A=2 cos2 A2 sin A cos A=1+cos 2Asin 2A Putting A=71∘2⇒cot 71∘2=1+cos 15∘sin 15∘ On simplification, We get cot 71∘2=√6+√2+√3+√4.

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to