Question

Prove that:
\[\dfrac{{\cos A\cos ecA - \sin A.\sec A}}{{\cos A + \sin A}} = \cos ecA - \sec A\]

Answer 1

Hint: Here we will solve the given equation by simplifying the L.H.S using the trigonometric formulae.

Complete step-by-step answer:
We have to prove \[\dfrac{{\cos A\cos ecA - \sin A.\sec A}}{{\cos A + \sin A}} = \cos ecA - \sec A\]. So let’s start simplifying the L.H.S part. Now $\cos ecA = \dfrac{1}{{\sin A}}$and $secA = \dfrac{1}{{\cos A}}$
Using it in the L.H.S part, we get
$\Rightarrow$ $\dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\cos A}}}}{{\cos A + \sin A}}$
Now taking LCM in the numerator part we get
$\Rightarrow$ $\dfrac{{{{\cos }^2}A - {{\sin }^2}A}}{{(\cos A + \sin A)(\sin A.\cos A)}}$
Now $({a^2} - {b^2}) = (a + b)(a - b)$ using this in numerator
$\Rightarrow$ $\dfrac{{(\cos A - \sin A)(\cos A + \sin A)}}{{(\sin A\cos A)(\sin A + \cos A)}} = \dfrac{{(\cos A - \sin A)}}{{\sin A.\cos A}}$
Segregating the denominator for separate terms in numerator we have
$\Rightarrow$ $\dfrac{{\cos A}}{{\sin A.\cos A}} - \dfrac{{\sin A}}{{\sin A.\cos A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\cos A}}$
$\Rightarrow$ $\cos ecA - \sec A$ = R.H.S.
Clearly, L.H.S=R.H.S.
Hence proved.

Note: Whenever you come across such questions, always start simplifying one side either L.H.S or R.H.S, using some trigonometric identities we can ultimately arrive at the required result.