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

Hint: 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:-
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
\[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]