Question

# If cosec A + sec A = cosec B + sec B, prove that $\tan A.\tan B=\cot \dfrac{A+B}{2}$

Hint: The given equation is in terms of cosec x and cot x. So, we must use any of the identities including them. To make the equation look simpler, there are many interrelations between cosec x, cot x, cos x, and sin x. As, $\theta \in R$, we can use the relation of sin x instead of cosec x and sin x, cos x instead of cot x. So, first, convert the whole equation in terms of cos x and sin x. Then, use any of the identities which makes the equation easy to solve. Here, use
$\tan x=\dfrac{\sin x}{\cos x};\cot x=\dfrac{\cos x}{\sin x};\operatorname{cosec}x=\dfrac{1}{\sin x};\sec x=\dfrac{1}{\cos x}$
$\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$
$\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$

Equality with sine, cosine, or tangent in them is called trigonometric equality. These are solved by some interrelations known beforehand. All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for proof. These are the main and crucial steps to take us nearer to the result.
Let us consider our question.
cosec A + sec A = cosec B + sec B
By general trigonometric knowledge, we know that
$\operatorname{cosec}A=\dfrac{1}{\sin A};\sec A=\dfrac{1}{\cos A}$
By substituting these values into our original equation, we get,
$\dfrac{1}{\sin A}+\dfrac{1}{\cos A}=\dfrac{1}{\sin B}+\dfrac{1}{\cos B}$
By subtracting $\dfrac{1}{\sin B}$ on both the sides, we get the equation as,
$\dfrac{1}{\sin A}-\dfrac{1}{\sin B}+\dfrac{1}{\cos A}=\dfrac{1}{\cos B}$
By subtracting $\dfrac{1}{\cos A}$ on both the sides, we get the equation as,
$\dfrac{1}{\sin A}-\dfrac{1}{\sin B}=\dfrac{1}{\cos B}-\dfrac{1}{\cos A}$
By taking the least common multiple on both the sides, we get,
$\dfrac{\sin B-\sin A}{\sin A\sin B}=\dfrac{\cos A-\cos B}{\cos A\cos B}$
By doing cross-multiplication, we get an equation of the form,
$\tan A\tan B=\dfrac{\sin B-\sin A}{\cos A-\cos B}$
By basic trigonometric knowledge, we get them as:
$\sin A-\sin B=2\sin \left( \dfrac{A-B}{2} \right)\cos \left( \dfrac{A+B}{2} \right)$
$\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
By taking “ – “ inside, we can write the formula as:
$\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$
By substituting these equations, into our equation, we get,
$\dfrac{\sin A\sin B}{\cos A\cos B}=\dfrac{2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)}{2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)}$
By canceling the common terms on the right-hand side, we get,
$\dfrac{\sin A}{\cos A}.\dfrac{\sin B}{\cos B}=\dfrac{\cos \left( \dfrac{A+B}{2} \right)}{\sin \left( \dfrac{A+B}{2} \right)}$
By basic knowledge of trigonometry, we know the relations as:
$\tan x=\dfrac{\sin x}{cox};\cot x=\dfrac{\cos x}{\sin x}$
By substituting these into our equation, we get it as:
$\tan A.\tan B=\cot \left( \dfrac{A+B}{2} \right)$
Hence proved.
Therefore, we have proved the required equation by the given condition.

Note: If you do the least common multiple without bringing sin, cos terms together, you cannot generate the term tan A. tan B. To generate that term, we must combine the terms. You need to be careful while substituting cos A – cos B, you must remove the negative sign only if you write B – A or else you must take the negative sign. Even if you solve by taking the ‘ – ‘ sign you will end up with the same result.