Question

Prove that
\[\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}\] \[=\] \[\dfrac{1}{\sin \left( a-b \right)}\left\{ \tan (x-b)-\tan (x-a) \right\}\]

Answer 1

Hint: we have to know the expansion of \[\sin (x-y)=\sin x\cos y-\cos x\sin y\]and we have to get thought that we have to multiply and divide with \[\sin \left( a-b \right)\] on both numerator and denominator by seeing the question.

Complete step-by-step answer:
First take left hand side \[\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}\]. . . . . . . . . . . . . . . . . . (1)
Now multiply with \[\sin \left( a-b \right)\] on both numerator and denominator that is equation (1) and we will get the expression as follows,
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right\}\]. . . . . . . . . . . . . . . (2)
Now in the numerator of \[\sin \left( a-b \right)\] add and subtract x so it makes no change in numerator that is in equation (2)
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin \left( a-x+x-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right\}\]. . . . . . . . . . . . . . . . (3)
Equation 3 can also be written as the below
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin ((x-b)-(x-a))}{\cos \left( x-b \right)\cos \left( x-b \right)} \right\}\]. . . . . . . . . . . . . . . . . . (4)
Now expand the numerator of the equation (4) we will get as follows,
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin (x-b)\cos (x-a)-\cos (x-b)\sin (x-a)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right\}\]. . . . . . . . . . . . . . . . . . . (5)
Separating the terms of numerator with denominator of equation (6) we will get,
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin (x-b)\cos (x-a)}{\cos \left( x-a \right)\cos \left( x-b \right)}-\dfrac{\cos (x-b)\sin (x-a)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right\}\]
Now cancelling the terms \[\cos \left( x-a \right)\]in the first part of fraction in both numerator and denominator and cancelling the terms \[\cos \left( x-b \right)\]in the second part of fraction in both numerator and denominator we will obtain the expression as
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \dfrac{\sin (x-b)}{\cos (x-b)}-\dfrac{\sin (x-a)}{\cos (x-a)} \right\}\]
\[\dfrac{1}{\sin \left( a-b \right)}\left\{ \tan (x-b)-\tan (x-a) \right\}\]
Hence proved that the left hand side is equal to the right hand side.

Note: we have to be clear that we have to get the required expression in the right hand side by doing some mathematical operations in trigonometry. In the trigonometry we have to do some operations to prove that expression in right hand side and that we have to think and comes by practice