Question

If tan A + tan B = p ; cot A + cot B = q ; cot ( A + B ) = ?

Answer 1

Answer:

Step-by-step explanation:

Her u get...

Hope u understand...

Answer 2

Given :

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

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

To find :

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

Solution :

Given that,

$\longrightarrow\sf{tan\;A+tan\;B=p}$

$\longrightarrow\sf{\dfrac{1}{cot\;A}+\dfrac{1}{cot\;B}=p}$

$\longrightarrow \sf{\dfrac{cot\;B+cot\;A}{cot\;A\;cot\;B}=p}$

using equation (1)

$\longrightarrow \sf{\dfrac{q}{cot\;A\;cot\;B}=p}$

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

Now,

using trigonometric identity

$\pink{\star}\;\;\boxed{\sf{cot(x+y)=\dfrac{cot\;x\;cot\;y-1}{cot\;x+cot\;y}}}$

$\longrightarrow\sf{\cot(A+B)=\dfrac{cot\;A\;cot\;B-1}{cot\;A+cot\;B}}$

using equation (1) and (2)

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

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

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

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

More related trigonometric identites

$\implies\sf{cos(x+y)=cos\;x\;cos\;y-sin\;x\;sin\;y}$

$\implies\sf{cos(x-y)=cos\;x\;cos\;y+sin\;x\;sin\;y}$

$\implies\sf{sin(x+y)=sin\;x\;cos\;y+cos\;x\;sin\;y}$

$\implies\sf{sin(x-y)=sin\;x\;cos\;y-cos\;x\;sin\;y}$

$\implies\sf{tan(x+y)=\dfrac{tan\;x+tan\;y}{1-tan\;x\;tan\;y}}$

$\implies\sf{tan(x-y)=\dfrac{tan\;x-tan\;y}{1+tan\;x\;tan\;y}}$

$\sf{cot(x-y)=\dfrac{cot\;x\;cot\;y+1}{cot\;y-cot\;x}}$