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Question: If we have a trigonometric expression as \(\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B\), pro...

If we have a trigonometric expression as cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B, prove that: tan⁡Atan⁡B=cot⁡A+B2\tan A\tan B=\cot \dfrac{A+B}{2}

Explanation

Solution

Here, we have been given that cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B and we have to prove that tan⁡Atan⁡B=cot⁡A+B2\tan A\tan B=\cot \dfrac{A+B}{2}. For this, we will first change cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B in the form of sin and cos by using cos⁡ecx=1sin⁡x\cos \text{ec}x=\dfrac{1}{\sin x} and sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}. Then we will solve and cross multiply then the obtained equation so that it comes in linear form. Then we will separate the terms with sinAsinB and cosAcosB on different sides of the equal to sign. Then we will take both these in common so that we will get the difference of angles in sin and cos. Then we will use the formulas sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C-\sin D=2\cos \dfrac{C+D}{2}\sin \dfrac{C-D}{2} and cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2} in the then obtained equation and then convert it in the form of tan and cot. Hence, we will get the required condition.

Complete step-by-step solution
Here, we have been given that cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec Band we have to prove that tan⁡Atan⁡B=cot⁡A+B2\tan A\tan B=\cot \dfrac{A+B}{2}. For this, we convert cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B in the form of sin and cos function first and then tan and cot and check if we get the required answer or not.
Now, we have:
cosecA+sec⁡A=cos⁡ecB+sec⁡B\text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B
Now we know that cos⁡ecx=1sin⁡x\cos \text{ec}x=\dfrac{1}{\sin x} and sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}.
Thus, we get the above equation as:
cosecA+sec⁡A=cos⁡ecB+sec⁡B ⇒1sin⁡A+1cos⁡A=1sin⁡B+1cos⁡B \begin{aligned} & \text{cosec}A+\sec A=\cos \text{ec}B\text{+}\sec B \\\ & \Rightarrow \dfrac{1}{\sin A}+\dfrac{1}{\cos A}=\dfrac{1}{\sin B}+\dfrac{1}{\cos B} \\\ \end{aligned}
Now, solving it, we get:
1sin⁡A+1cos⁡A=1sin⁡B+1cos⁡B ⇒cos⁡A+sin⁡Asin⁡Acos⁡A=cos⁡B+sin⁡Bsin⁡Bcos⁡B \begin{aligned} & \dfrac{1}{\sin A}+\dfrac{1}{\cos A}=\dfrac{1}{\sin B}+\dfrac{1}{\cos B} \\\ & \Rightarrow \dfrac{\cos A+\sin A}{\sin A\cos A}=\dfrac{\cos B+\sin B}{\sin B\cos B} \\\ \end{aligned}
Now, cross multiplying the above obtained equation, we get:
cos⁡A+sin⁡Asin⁡Acos⁡A=cos⁡B+sin⁡Bsin⁡Bcos⁡B ⇒(cos⁡A+sin⁡A)sin⁡Bcos⁡B=(cos⁡B+sin⁡B)sin⁡Acos⁡A ⇒cos⁡Asin⁡Bcos⁡B+sin⁡Asin⁡Bcos⁡B=sin⁡Acos⁡Acos⁡B+sin⁡Acos⁡Asin⁡B \begin{aligned} & \dfrac{\cos A+\sin A}{\sin A\cos A}=\dfrac{\cos B+\sin B}{\sin B\cos B} \\\ & \Rightarrow \left( \cos A+\sin A \right)\sin B\cos B=\left( \cos B+\sin B \right)\sin A\cos A \\\ & \Rightarrow \cos A\sin B\cos B+\sin A\sin B\cos B=\sin A\cos A\cos B+\sin A\cos A\sin B \\\ \end{aligned}
Now, we can also write it as:

& \cos A\sin B\cos B+\sin A\sin B\cos B=\sin A\cos A\cos B+\sin A\cos A\sin B \\\ & \Rightarrow \cos A\sin B\cos B-\sin A\cos A\cos B=\sin A\cos A\sin B-\sin A\sin B\cos B \\\ \end{aligned}$$ Now, taking cosAcosB common in the LHS and sinAsinB common in the RHS we get: $$\begin{aligned} & \cos A\sin B\cos B-\sin A\cos A\cos B=\sin A\cos A\sin B-\sin A\sin B\cos B \\\ & \Rightarrow \cos A\cos B\left( \sin B-\sin A \right)=\sin A\sin B\left( \cos A-\cos B \right) \\\ \end{aligned}$$ Now, we know that $\sin C-\sin D=2\cos \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ and $\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ Hence, we get the above obtained equation as: $$\begin{aligned} & \cos A\cos B\left( \sin B-\sin A \right)=\sin A\sin B\left( \cos A-\cos B \right) \\\ & \Rightarrow \cos A\cos B\left( 2\cos \dfrac{B+A}{2}\sin \dfrac{B-A}{2} \right)=\sin A\sin B\left( -2\sin \dfrac{A+B}{2}\sin \dfrac{A-B}{2} \right) \\\ \end{aligned}$$ Now, solving this equation we get: $$\begin{aligned} & \cos A\cos B\left( 2\cos \dfrac{B+A}{2}\sin \dfrac{B-A}{2} \right)=\sin A\sin B\left( -2\sin \dfrac{A+B}{2}\sin \dfrac{A-B}{2} \right) \\\ & \Rightarrow \cos A\cos B\left( 2\cos \dfrac{B+A}{2}\sin \dfrac{B-A}{2} \right)=\sin A\sin B\left( 2\sin \dfrac{A+B}{2}\sin \dfrac{B-A}{2} \right) \\\ & \Rightarrow \cos A\cos B\cos \dfrac{A+B}{2}=\sin A\sin B\sin \dfrac{A+B}{2} \\\ & \Rightarrow \dfrac{\cos \dfrac{A+B}{2}}{\sin \dfrac{A+B}{2}}=\dfrac{\sin A\sin B}{\cos A\cos B} \\\ \end{aligned}$$ Hence, we get: $$\begin{aligned} & \dfrac{\cos \dfrac{A+B}{2}}{\sin \dfrac{A+B}{2}}=\dfrac{\sin A\sin B}{\cos A\cos B} \\\ & \therefore \cot \dfrac{A+B}{2}=\tan A\tan B \\\ \end{aligned}$$ Hence, the given condition in the question is proved. **Note:** Some formulas as the sum or difference of sin and cos are given below which might come in handy: 1\. $\sin C+\sin D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$ 2\. $\sin C-\sin D=2\cos \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ 3\. $\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$ 4\. $\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$