Question

Prove that: \[(\text{cosec}A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1\]

Answer 1

Hint: We will begin with the left hand side of the equation (1) and then we will solve it by first converting all the terms in sin and cos and then we will take the LCM and finally we will get the answer which will be equal to the right hand side of the expression. We will use the formula \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] which will help us in proving the given expression.

Complete step-by-step answer:

It is mentioned in the question that \[(\text{cosec}A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1......(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow (\text{cosec}A-\sin A)(\sec A-\cos A)(\tan A+\cot A).....(2)\]

Now converting sec, cosec, tan and cot in equation (2) in terms of cos and sin and hence we get,

\[\Rightarrow \left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right).....(3)\]

Taking the LCM in equation (3) and simplifying we get,

\[\Rightarrow \left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A}

\right)\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right).......(4)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (4) we get,

\[\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A}

\right)\left( \dfrac{1}{\sin A\cos A} \right).......(5)\]

Now cancelling similar terms in the numerator and denominator in equation (5) we get.

\[\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A}
\right)\left( \dfrac{1}{\sin A\cos A} \right)=1.......(6)\]

Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (4) but here the key is to substitute 1 in place of \[{{\sin }^{2}}A+{{\cos }^{2}}A\]. Then we need to divide each term in the numerator separately by the denominator in equation (5) to get the left hand side equal to the right hand side of the equation (1).