Question

Prove that:
(i) $\dfrac{{\cos A}}{{1 - \sin A}} = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$
(ii) $\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {80^ \circ } = \dfrac{3}{{16}}$

Answer 1

Here we will use various identities and values of certain trigonometric ratios.
The identities we will use are:-
$\sin 2\theta = 2\sin \theta \cos \theta$
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
${\cos ^2}\theta + {\sin ^2}\theta = 1$
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\sin A\sin \left( {{{60}^ \circ } - A} \right)\sin \left( {{{60}^ \circ } + A} \right) = \dfrac{1}{4}\sin 3A$

Complete step-by-step answer:
Let us first consider part (i)
(i) $\dfrac{{\cos A}}{{1 - \sin A}} = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$
Let us consider the left hand side we get:
$LHS = \dfrac{{\cos A}}{{1 - \sin A}}$………………………. (1)
Now we know that:-
$\sin 2\theta = 2\sin \theta \cos \theta$
Hence, $\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$
Also, we know that:-
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
Hence, $\cos A = {\cos ^2}\dfrac{A}{2} - {\sin ^2}\dfrac{A}{2}$
Putting these values in equation 1 we get:-
$LHS = \dfrac{{{{\cos }^2}\dfrac{A}{2} - {{\sin }^2}\dfrac{A}{2}}}{{1 - 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}$
Now we know that:-
${\cos ^2}\theta + {\sin ^2}\theta = 1$
Hence,
${\cos ^2}\dfrac{A}{2} + {\sin ^2}\dfrac{A}{2} = 1$
Substituting this value in the above equation we get:-
$LHS = \dfrac{{{{\cos }^2}\dfrac{A}{2} - {{\sin }^2}\dfrac{A}{2}}}{{{{\cos }^2}\dfrac{A}{2} + {{\sin }^2}\dfrac{A}{2} - 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}$
Now we know that:-
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Applying this identity in the denominator we get:-
$LHS = \dfrac{{{{\cos }^2}\dfrac{A}{2} - {{\sin }^2}\dfrac{A}{2}}}{{{{\left( {\cos \dfrac{A}{2} - \sin \dfrac{A}{2}} \right)}^2}}}$
Now applying the following identity in the numerator:-
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
We get:-
$LHS = \dfrac{{\left( {\cos \dfrac{A}{2} + \sin \dfrac{A}{2}} \right)\left( {\cos \dfrac{A}{2} - \sin \dfrac{A}{2}} \right)}}{{{{\left( {\cos \dfrac{A}{2} - \sin \dfrac{A}{2}} \right)}^2}}}$
Cancelling the terms we get:-
$LHS = \dfrac{{\left( {\cos \dfrac{A}{2} + \sin \dfrac{A}{2}} \right)}}{{\left( {\cos \dfrac{A}{2} - \sin \dfrac{A}{2}} \right)}}$
Now dividing the numerator and the denominator by $\cos \dfrac{A}{2}$ we get:-
$LHS = \dfrac{{\left( {\dfrac{{\cos \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} + \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}} \right)}}{{\left( {\dfrac{{\cos \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}} - \dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}} \right)}}$
Simplify it further we get:-
$LHS = \dfrac{{\left( {1 + \tan \dfrac{A}{2}} \right)}}{{\left( {1 - \left( 1 \right)\tan \dfrac{A}{2}} \right)}}$
Now we know that:-
$\tan \dfrac{\pi }{4} = 1$
Substituting this value above equation we get:-
$LHS = \dfrac{{\left( {\tan \dfrac{\pi }{4} + \tan \dfrac{A}{2}} \right)}}{{\left( {1 - \tan \dfrac{\pi }{4}\tan \dfrac{A}{2}} \right)}}$
Now we know that:-
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Hence, applying this identity we get:-
$LHS = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$
Also, $RHS = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$
Therefore, $LHS = RHS$
Hence proved.
(ii) $\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {80^ \circ } = \dfrac{3}{{16}}$
Let us consider the left hand side:-
$LHS = \sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {80^ \circ }$
$\Rightarrow LHS = \sin {60^ \circ }\left[ {\sin {{20}^ \circ }\sin {{40}^ \circ }\sin {{80}^ \circ }} \right]$
Now we know that:-
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\sin {40^ \circ } = \sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)$
$\sin {80^ \circ } = \sin \left( {{{60}^ \circ } + {{20}^ \circ }} \right)$
Hence substituting these values we get:-
$LHS = \dfrac{{\sqrt 3 }}{2}\left[ {\sin {{20}^ \circ }\sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)\sin \left( {{{60}^ \circ } + {{20}^ \circ }} \right)} \right]$
Now we know that:-
$\sin A\sin \left( {{{60}^ \circ } - A} \right)\sin \left( {{{60}^ \circ } + A} \right) = \dfrac{1}{4}\sin 3A$
Applying this identity in above equation we get:-
$LHS = \dfrac{{\sqrt 3 }}{2}\left[ {\dfrac{1}{4}\sin 3\left( {{{20}^ \circ }} \right)} \right]$
Simplifying it further we get:-
$LHS = \dfrac{{\sqrt 3 }}{8}\left[ {\sin {{60}^ \circ }} \right]$
We know that:-
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Putting the value we get:-
$LHS = \dfrac{{\sqrt 3 }}{8}\left[ {\dfrac{{\sqrt 3 }}{2}} \right]$
Simplifying it we get:-
$LHS = \dfrac{3}{{16}}$
Now, since $RHS = \dfrac{3}{{16}}$
Hence, $LHS = RHS$
Hence proved.

Note: In part (ii) students can also, use the following identities to solve $\dfrac{{\sqrt 3 }}{2}\left[ {\sin {{20}^ \circ }\sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right)\sin \left( {{{60}^ \circ } + {{20}^ \circ }} \right)} \right]$but it would be a bit lengthy and tedious as we would have to evaluate the value of $\sin {20^ \circ }$
The formulas are:-
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$