Prove that following identities-A -60- A -60- A -3 3 A

LHS = cot A + cot (60^∘ + A) cot (60^∘ A) = 1/tan A+1/tan (60^∘ + A) 1/tan 60^∘ A=1/tan A+1 √(3) tan A/√(3)+tan A 1+√(3) tan A/√(3) tan A = 1/tan A 8 tan A ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove that fo...

Question

Prove that following identities:

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

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Solution

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

$= \frac{1}{t a n A} + \frac{1}{t a n (60^{\circ} + A)} - \frac{1}{t a n 60^{\circ} - A} = \frac{1}{t a n A} + \frac{1 - \sqrt{3} t a n A}{\sqrt{3} + t a n A} - \frac{1 + \sqrt{3} t a n A}{\sqrt{3} - t a n A} = \frac{1}{t a n A} - \frac{8 t a n A}{3 - t a n^{2} A} = \frac{3 - t a n^{2} A - 8 t a n^{2} A}{3 t a n A - t a n^{3} A} = \frac{3 - 9 t a n^{2} A}{3 t a n A - t a n^{3} A} = 3 (\frac{1 - 3 t a n^{2} A}{3 t a n A - t a n^{3} A}) = \frac{3}{t a n 3 A} = 3 c o t 3 A = R H S$
LHS = RHS

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Q. 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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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) - c o t (60 ° - A) = 3 \cot 3 A

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

Q. Prove the following identities:

|\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}| = a^{2} (a + x + y + z)

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Prove that following identities: cot A+cot (60∘+A)−cot (60∘−A)=3 cot 3A

LHS=cot A+cot (60∘+A)−cot (60∘−A) =1tan A+1tan (60∘+A)−1tan 60∘−A=1tan A+1−√3 tan A√3+tan A−1+√3 tan A√3−tan A=1tan A−8 tan A3−tan2 A=3−tan2 A−8 tan2 A3tanA−tan3 A=3−9 tan2 A3 tan A−tan3 A=3(1−3 tan2 A3 tan A−tan3A)=3tan 3A=3 cot 3A=RHS LHS = RHS

Prove that following identities:

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