Questions & Answers

Question

Answers

$(\sec A - \csc A)(1 + \tan A + \cot A) = \tan A\sec A - \cot A\csc A.$

Answer
Verified

Hint-To prove this question, we will use basic trigonometric identities such as

$

{\sin ^2}\theta + {\cos ^2}\theta = 1,\sin \theta = \dfrac{1}{{\csc \theta }},\cos \theta = \dfrac{1}{{\sec \theta }} \\

\tan \theta = \dfrac{1}{{\cot \theta }} = \dfrac{{\sin \theta }}{{\cos \theta }} \\

$

Complete step-by-step solution -

Given expression is

$(\sec A - \csc A)(1 + \tan A + \cot A) = \tan A\sec A - \cot A\csc A$

To proceed further we will use L.H.S of the above equation

Substituting the values of \[\sec A,\csc A,\tan A\] in terms of $\sin {\text{A and cosA}}$$ = \left[ {\dfrac{1}{{\cos A}} - \dfrac{1}{{\sin A}}} \right]\left[ {1 + \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\cos A}}{{\sin A}}} \right]$

Simplifying the above equation further, we get

$

= \left[ {\dfrac{{\sin A - \cos A}}{{\sin A\cos A}}} \right]\left[ {\dfrac{{\cos A\sin A + {{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}}} \right] \\

= \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{{{\sin }^2}A{{\cos }^2}A}} \\

$

As we know that $\left[ {{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})} \right]$

$ = \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{{{\sin }^2}A{{\cos }^2}A}}......................(2)$

Now, we will take the R.H.S, of the given expression

\[ = \tan A\sec A - \cot A\csc A\]

Writing the above equation in terms of $\sin A,\cos A$

$

= \dfrac{{\sin A}}{{\cos A}} \times \dfrac{1}{{\cos A}} - \dfrac{{\cos A}}{{\sin A}} \times \dfrac{1}{{\sin A}} \\

= \dfrac{{\sin A}}{{{{\cos }^2}A}} - \dfrac{{\cos A}}{{{{\sin }^2}A}} \\

= \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{{{\cos }^2}A{{\sin }^2}A}}...........................(3) \\

$

From equation 2 and 3 it is clear that the L.H.S=R.H.S

Note- In order to solve such questions, remember the basic trigonometric identities which are also listed above. Always try to manipulate the equations keeping in mind the RHS or the equation to be proved, as sometimes there are multiple manipulations available in trigonometry but not all are required. Try to use algebraic formulas when necessary in order to solve the equation quickly.

$

{\sin ^2}\theta + {\cos ^2}\theta = 1,\sin \theta = \dfrac{1}{{\csc \theta }},\cos \theta = \dfrac{1}{{\sec \theta }} \\

\tan \theta = \dfrac{1}{{\cot \theta }} = \dfrac{{\sin \theta }}{{\cos \theta }} \\

$

Complete step-by-step solution -

Given expression is

$(\sec A - \csc A)(1 + \tan A + \cot A) = \tan A\sec A - \cot A\csc A$

To proceed further we will use L.H.S of the above equation

Substituting the values of \[\sec A,\csc A,\tan A\] in terms of $\sin {\text{A and cosA}}$$ = \left[ {\dfrac{1}{{\cos A}} - \dfrac{1}{{\sin A}}} \right]\left[ {1 + \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\cos A}}{{\sin A}}} \right]$

Simplifying the above equation further, we get

$

= \left[ {\dfrac{{\sin A - \cos A}}{{\sin A\cos A}}} \right]\left[ {\dfrac{{\cos A\sin A + {{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}}} \right] \\

= \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{{{\sin }^2}A{{\cos }^2}A}} \\

$

As we know that $\left[ {{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})} \right]$

$ = \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{{{\sin }^2}A{{\cos }^2}A}}......................(2)$

Now, we will take the R.H.S, of the given expression

\[ = \tan A\sec A - \cot A\csc A\]

Writing the above equation in terms of $\sin A,\cos A$

$

= \dfrac{{\sin A}}{{\cos A}} \times \dfrac{1}{{\cos A}} - \dfrac{{\cos A}}{{\sin A}} \times \dfrac{1}{{\sin A}} \\

= \dfrac{{\sin A}}{{{{\cos }^2}A}} - \dfrac{{\cos A}}{{{{\sin }^2}A}} \\

= \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{{{\cos }^2}A{{\sin }^2}A}}...........................(3) \\

$

From equation 2 and 3 it is clear that the L.H.S=R.H.S

Note- In order to solve such questions, remember the basic trigonometric identities which are also listed above. Always try to manipulate the equations keeping in mind the RHS or the equation to be proved, as sometimes there are multiple manipulations available in trigonometry but not all are required. Try to use algebraic formulas when necessary in order to solve the equation quickly.

×

Sorry!, This page is not available for now to bookmark.