How can we prove A- 60- A - 60- A -3 3 A

Cot A+cot(60+a) cot(60 a)=3 cot 3a =cot A + cos(60+a)/sin(60+a) cos(60 a)/sin(60 a) =cot A+cos A √(3)sin A/√(3)cos A+sin A cos A+√(3)sin A/√(3)cos A sin A ⇒ cot ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

How can we pr...

Question

How can we prove $c o t A + c o t (60 + A) + c o t (60 - A) = 3 c o t 3 A$

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Solution

$c o t A + c o t (60 + a) - c o t (60 - a) = 3 c o t 3 a$
$= c o t A + \frac{c o s (60 + a)}{s i n (60 + a)} - \frac{c o s (60 - a)}{s i n (60 - a)}$
$= c o t A + \frac{c o s A - \sqrt{3} s i n A}{\sqrt{3} c o s A + s i n A} - \frac{c o s A + \sqrt{3} s i n A}{\sqrt{3} c o s A - s i n A}$
$\Rightarrow c o t A + \frac{- 8 s i n A c o s A}{3 c o s^{2} A - s i n^{2} A}$
$\Rightarrow c o t A - \frac{8 c o t A}{3 c o t^{2} A - 1}$
$c o t A + c o t (60 + a) - c o t (60 - a) = 3 c o t 3 a$
$\Rightarrow \frac{3 c o t^{3} A - 9 c o t A}{3 c o t^{2} A - 1}$
$\Rightarrow 3 (\frac{c o t^{3} A - 3 c o t A}{3 c o t^{2} A - 1})$
$\Rightarrow 3 c o t 3 A$

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Similar questions

Prove that following identities:

$c o t A + c o t (60^{\circ} + A) - c o t (60^{\circ} - A) = 3 c o t 3 A$

\cot A + \cot (60 ° + A) - c o t (60 ° - A) = 3 \cot 3 A

Provie that $s i n A s i n (60^{\circ} - A) s i n (60^{\circ} + A) = \frac{4}{4} s i n 3 A .$

Hene , deduce that $s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s o m 80^{\circ} = \frac{3}{16} .$

Prove tahat $c o t A + c o t (60^{\circ} + A) - c o t (60^{\circ} - A) = 3 c o t 3 A .$

\cot A + \cot (60 ° + A) + \cot (120 ° + A) = 3 \cot 3 A

We can construct a $60^{\circ}$ angle with a compass and ruler.

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How can we prove cotA+cot(60+A)+cot(60−A)=3 cot 3A

How can we prove $c o t A + c o t (60 + A) + c o t (60 - A) = 3 c o t 3 A$