Question

if a = tan 25 + tan 35 - root3 and b = cot 25 + cot 35 + root 3 then ab =?

Answer 1

^^^^^^^^^^^^^^^^^^^↑^^^^^^^^^^^^^^^^^^^^^

Answer 2

$\mathsf{ab=-3}$

Concept to be used:

$\mathsf{tan(A+B)=\dfrac{tanA+tanB}{1-tanA\:tanB}}$

$\mathsf{cot(A+B)=\dfrac{cotA\:cotB-1}{cotB+cotA}}$

$\mathsf{tanA\times cotA=1}$

Step-by-step explanation:

Step 1.

We know that, $\mathsf{tan60^{\circ}=\sqrt{3}}$

$\mathsf{\Rightarrow tan(25^{\circ}+35^{\circ})=\sqrt{3}}$

$\mathsf{\Rightarrow \dfrac{tan25^{\circ}+tan35^{\circ}}{1-tan25^{\circ}\:tan35^{\circ}}=\sqrt{3}}$

$\mathsf{\Rightarrow tan25^{\circ}+tan35^{\circ}=\sqrt{3}-\sqrt{3}\:tan25^{\circ}\:tan35^{\circ}}$

$\mathsf{\Rightarrow tan25^{\circ}+tan35^{\circ}-\sqrt{3}=-\sqrt{3}\:tan25^{\circ}\:tan35^{\circ}}$

So, we have

$\boxed{\mathsf{a=-\sqrt{3}\:tan25^{\circ}\:tan35^{\circ}}}$

Step 2.

Also, $\mathsf{cot60^{\circ}=\dfrac{1}{\sqrt{3}}}$

$\mathsf{\Rightarrow cot(25^{\circ}+35^{\circ})=\dfrac{1}{\sqrt{3}}}$

$\mathsf{\Rightarrow \dfrac{cot25^{\circ}\:cot35^{\circ}-1}{cot25^{\circ}+cot35^{\circ}}=\dfrac{1}{\sqrt{3}}}$

$\mathsf{\Rightarrow \sqrt{3}\:cot25^{\circ}\:cot35^{\circ}-\sqrt{3}=cot25^{\circ}+cot35^{\circ}}$

$\mathsf{\Rightarrow cot25^{\circ}+cot35^{\circ}+\sqrt{3}=\sqrt{3}\:cot25^{\circ}\:cot35^{\circ}}$

So, we have

$\boxed{\mathsf{b=\sqrt{3}\:cot25^{\circ}\:cot35^{\circ}}}$

Step 3.

Now, $\mathsf{ab}$

$\mathsf{=(-\sqrt{3}\:tan25^{\circ}\:tan35^{\circ})(\sqrt{3}\:cot25^{\circ}\:cot35^{\circ})}$

$\mathsf{=-\sqrt{3}\times\sqrt{3}\times tan25^{\circ}\times cot25^{\circ}\times tan35^{\circ}\times cot35^{\circ}}$

$\mathsf{=-3\times 1\times 1}$

$\mathsf{=-3}$

#SPJ3