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Question: Prove that \[\dfrac{\tan A}{\sec A+1}+\dfrac{\cot A}{\text{cosec}A+1}=\text{cosec}A+\sec A-\sec A\ta...

Prove that tan⁡Asec⁡A+1+cot⁡AcosecA+1=cosecA+sec⁡A−sec⁡Atan⁡A\dfrac{\tan A}{\sec A+1}+\dfrac{\cot A}{\text{cosec}A+1}=\text{cosec}A+\sec A-\sec A\tan A

Explanation

Solution

Hint : In order to prove tan⁡Asec⁡A+1+cot⁡AcosecA+1=cosecA+sec⁡A−sec⁡Atan⁡A\dfrac{\tan A}{\sec A+1}+\dfrac{\cot A}{\text{cosec}A+1}=\text{cosec}A+\sec A-\sec A\tan A, we will be considering the LHS first and then we will be rationalizing the denominators. After rationalizing, we will solve them accordingly with the help of trigonometric ratios by cancelling the common terms and then considering the common terms and upon solving it, we will be able to prove that LHs is equal to RHS and that would be our required solution.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between 0∘{{0}^{\circ }} and 360∘{{360}^{\circ }}. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us start proving tan⁡Asec⁡A+1+cot⁡AcosecA+1=cosecA+sec⁡A−sec⁡Atan⁡A\dfrac{\tan A}{\sec A+1}+\dfrac{\cot A}{\text{cosec}A+1}=\text{cosec}A+\sec A-\sec A\tan A
Firstly, let us consider the LHS, i.e. tan⁡Asec⁡A+1+cot⁡AcosecA+1\dfrac{\tan A}{\sec A+1}+\dfrac{\cot A}{\text{cosec}A+1}
Now let us rationalize the denominator. We get
⇒tan⁡A(sec⁡A−1)sec⁡A+1(sec⁡A−1)+cot⁡A(cosecA−1)cosecA+1(cosecA−1)\Rightarrow \dfrac{\tan A\left( \sec A-1 \right)}{\sec A+1\left( \sec A-1 \right)}+\dfrac{\cot A\left( \text{cosec}A-1 \right)}{\text{cosec}A+1\left( \text{cosec}A-1 \right)}
Upon solving this further, we get
⇒tan⁡A(sec⁡A−1)sec⁡2A−1+cot⁡A(cosecA−1)cosecA2−1\Rightarrow \dfrac{\tan A\left( \sec A-1 \right)}{{{\sec }^{2}}A-1}+\dfrac{\cot A\left( \text{cosec}A-1 \right)}{\text{cosec}{{A}^{2}}-1}
⇒tan⁡A(sec⁡A−1)tan⁡2A+cot⁡A(cosecA−1)cot⁡A2\Rightarrow \dfrac{\tan A\left( \sec A-1 \right)}{{{\tan }^{2}}A}+\dfrac{\cot A\left( \text{cosec}A-1 \right)}{\cot {{A}^{2}}}
Upon cancelling the common terms, we get
⇒(sec⁡A−1)tan⁡A+(cosecA−1)cot⁡A\Rightarrow \dfrac{\left( \sec A-1 \right)}{\tan A}+\dfrac{\left( \text{cosec}A-1 \right)}{\cot A}
Now upon splitting the terms, we get
⇒sec⁡Atan⁡A−1tan⁡A+cosecAcot⁡A−1cot⁡A\Rightarrow \dfrac{\sec A}{\tan A}-\dfrac{1}{\tan A}+\dfrac{\text{cosec}A}{\cot A}-\dfrac{1}{\cot A}
Upon simplifying the terms with respect to trigonometric ratios, we get
⇒cosecA+sec⁡A−[1tan⁡A+1cot⁡A]\Rightarrow \text{cosec}A+\sec A-\left[ \dfrac{1}{\tan A}+\dfrac{1}{\cot A} \right]
On further splitting the terms, we get
⇒cosecA+sec⁡A−[1sin⁡Acos⁡A+1cos⁡Asin⁡A]\Rightarrow \text{cosec}A+\sec A-\left[ \dfrac{1}{\dfrac{\sin A}{\cos A}}+\dfrac{1}{\dfrac{\cos A}{\sin A}} \right]
On solving this,
⇒cosecA+sec⁡A−[cos⁡Asin⁡A+sin⁡Acos⁡A]\Rightarrow \text{cosec}A+\sec A-\left[ \dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A} \right]

& \Rightarrow \text{cosec}A+\sec A-\left[ \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\cos A\sin A} \right] \\\ & \\\ \end{aligned}$$ We know that, $${{\cos }^{2}}A+{{\sin }^{2}}A=1$$. So we will be substituting this value in the expression, we get $$\Rightarrow \text{cosec}A+\sec A-\left[ \dfrac{1}{\cos A\sin A} \right]$$ $$\Rightarrow \text{cosec}A+\sec A-\sec A\text{cosec}A$$ Hence proved **Note:** While solving such problems, we must consider either LHS or RHS which would be easier in solving and proving. We must be aware of the trigonometric conversions perfectly in order to obtain the correct answer. In the above problem, we have only split the terms, taken the common terms and solved them.