Question

if tan a + tan b = p and cot a + cot b = q then cot (a + b ) = ?​

Answer 1

Given :

$\pink{\bullet}\;\rm{tan\;A+tan\;B=p}$

$\pink{\bullet}\;\rm{cot\;A+cot\;B=q\;\;\;...eqn(1)}$

To find :

$\pink{\bullet}\;\rm{cot(A+B) = ?}$

Solution :

Given that,

$\implies\rm{tan\;A+tan\;B=p}$

$\implies\rm{\dfrac{1}{cot\;A}+\dfrac{1}{cot\;B}=p}$

$\implies \rm{\dfrac{cot\;B+cot\;A}{cot\;A\;cot\;B}=p}$

using equation (1)

$\implies\rm{\dfrac{q}{cot\;A\;cot\;B}=p}$

$\implies \rm{cot\;A\;cot\;B=\dfrac{q}{p}\;\;\;....eqn(2)}$

Now,

using trigonometric identity

$\pink{\star}\;\; \boxed{\rm{cot(x+y)=\dfrac{cot\;x\;cot\;y-1}{cot\;x+cot\;y}}}</p><p>$

$\implies\rm{\cot(A+B)=\dfrac{cot\;A\;cot\;B-1}{cot\;A+cot\;B}}$

using equation (1) and (2)

$\implies\rm{\cot(A+B)=\dfrac{\dfrac{q}{p}-1}{q}}$

$\implies\rm{\cot(A+B)=\dfrac{q-p}{p}\div q}$

$\implies\rm{\cot(A+B)=\dfrac{q-p}{p}\times \dfrac{1}{q}}$

$\implies\overbrace{\underbrace{\boxed{\rm{\red{\cot(A+B)=\dfrac{q-p}{pq}}}}}}\;\;\;\;\pink{\bigstar}$

OTHER INFORMATION

TRIGNOMETRIC IDENTITIES

• sin²∅ + cos²∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

TRIGNOMETRIC RATIOS

• sin ∅ = 1 / cosec ∅

• cos ∅ = 1 / sec ∅

• tan ∅ = 1 / cot ∅

TRIGNOMETRIC COMPLEMENTRY ANGLES

• sin ∅ = cos ( 90 - ∅ )

• cos ∅ = sin ( 90 - ∅ )

• sec ∅ = cosec ( 90 - ∅ )

• cosec ∅ = sec ( 90 - ∅ )

• tan ∅ = cot ( 90 - ∅ )

• cot ∅ = tan ( 90 - ∅ )

Answer 2

Who is chandrasekhar azad?

Chandra Shekhar Azad ( sometimes also spelled Chandrasekhar; 23 July 1906 – 27 February 1931), popularly known as by his self-taken name Azad ("The Free"), was an Indian revolutionary who reorganised the Hindustan Republican Association under its new name of Hindustan Socialist Republican Army (HSRA) after the death of its founder, ...