Question

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

Answer 1

In this problem they give the trigonometric relation to prove. We are going to prove this problem by using the method, first consider the left hand side and from that we will derive the right hand side.

Formula used: We have used the following formula to derive the right hand side in i),
Algebraic formula,
${a^2} - {b^2} = (a + b)(a - b)$
${(a + b)^2} = {a^2} + 2ab + {b^2}$
Trigonometric formula,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\sin 2\theta = 2\sin \theta \cos \theta$
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
$\dfrac{{\sin {\text{ }}\theta }}{{\cos {\text{ }}\theta }} = \tan \;\theta$
$\tan \left( {a + b} \right) = \dfrac{{\tan {\text{ a + tan b}}}}{{1 - \left( {\tan {\text{ a}}} \right)\left( {\tan {\text{ b}}} \right)}}$
We have used the following formula to derive the right hand side in ii),
Trigonometric formula,
$2\sin x\sin y = \cos (x - y) - \cos (x + y)$
$2\sin x\cos y = \sin (x + y) + \sin (x - y)$
$\sin ( - \theta ) = - \sin \theta$

Complete step-by-step answer:
1. We need to prove $\dfrac{{\cos A}}{{1 - \sin A}} = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$.
Let us consider the L.H.S $= \dfrac{{\cos A}}{{1 - \sin A}}$
We will multiply and divide the same term $1 + \sin A$, we get
$\Rightarrow \dfrac{{\cos A}}{{1 - \sin A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}$
Simplifying we get,
$\Rightarrow \dfrac{{\cos A(1 + \sin A)}}{{(1 - \sin A)(1 + \sin A)}}$
Using the algebraic formula ${a^2} - {b^2} = (a + b)(a - b)$ for denominator,
$\Rightarrow \dfrac{{\cos A(1 + \sin A)}}{{(1 - {{\sin }^2}A)}}$
We know, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ by using this,
$\Rightarrow \dfrac{{\cos A(1 + \sin A)}}{{{{\cos }^2}A}}$
Simplifying we get,
$\Rightarrow \dfrac{{1 + \sin A}}{{\cos A}}$
By using the $\sin 2\theta = 2\sin \theta \cos \theta$ to get the required result,
$\Rightarrow \dfrac{{{{\sin }^2}\dfrac{A}{2} + {{\cos }^2}\dfrac{A}{2} + 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}{{{{\cos }^2}\dfrac{A}{2} - {{\sin }^2}\dfrac{A}{2}}}$
Now, we can use the algebraic formula, ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and the trigonometric formula,$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$

$\Rightarrow \dfrac{{{{(\sin \dfrac{A}{2} + \cos \dfrac{A}{2})}^2}}}{{(\cos \dfrac{A}{2} + \sin \dfrac{A}{2})(\cos \dfrac{A}{2} - \sin \dfrac{A}{2})}}$
Cancelling the common terms in numerator and denominator we get,
$\Rightarrow \dfrac{{\sin \dfrac{A}{2} + \cos \dfrac{A}{2}}}{{\cos \dfrac{A}{2} - \sin \dfrac{A}{2}}}$
Dividing denominator and numerator by $\cos \dfrac{A}{2}$,
$\Rightarrow \dfrac{{\dfrac{{\sin \dfrac{A}{2} + \cos \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}}}{{\dfrac{{\cos \dfrac{A}{2} - \sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}}}$
By using the formula $\dfrac{{\sin {\text{ }}\theta }}{{\cos {\text{ }}\theta }} = \tan \;\theta$ we get,
$\Rightarrow \dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{\pi }{4}}}{{1 - \tan \dfrac{A}{2}\tan \dfrac{\pi }{4}}}$
We know that is $\tan \left( {a + b} \right) = \dfrac{{\tan {\text{ a + tan b}}}}{{1 - \left( {\tan {\text{ a}}} \right)\left( {\tan {\text{ b}}} \right)}}$ which gives,
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$
=R.H.S.
Hence L.H.S = R.H.S.
$\therefore \dfrac{{\cos A}}{{1 - \sin A}} = \tan \left( {\dfrac{\pi }{4} + \dfrac{A}{2}} \right)$(proved).
2. We need to prove $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{{16}}$
L.H.S =
$\Rightarrow \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$
Multiply and divide by $2$ we get,
$\Rightarrow \dfrac{1}{2}(\sin 20^\circ \sin 60^\circ )(2\sin 40^\circ \sin 80^\circ )$
We know, $2\sin x\sin y = \cos (x - y) - \cos (x + y)$ by using,
$\Rightarrow \dfrac{1}{2}(\sin 20^\circ \sin 60^\circ )[\cos (80^\circ - 40^\circ ) - \cos (80^\circ + 40^\circ )]$
Substituting the value $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$,
$\Rightarrow \dfrac{1}{2}\left( {\sin 20^\circ \dfrac{{\sqrt 3 }}{2}} \right)\left( {\cos 40^\circ - \cos 120^\circ } \right)$
Simplifying we get,
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\cos 40^\circ + \dfrac{1}{2}} \right)$
Again we can use, $2\sin x\cos y = \sin (x + y) + \sin (x - y)$
$\Rightarrow \dfrac{{\sqrt 3 }}{8}2\sin 20^\circ \cos 40^\circ + \dfrac{{\sqrt 3 }}{8}\sin 20^\circ$
$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left[ {\sin (20^\circ + 40^\circ ) + \sin (20^\circ - 40^\circ )} \right] + \dfrac{{\sqrt 3 }}{8}\sin 20^\circ$
Simplifying we get,
$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left[ {\sin 60^\circ + \sin ( - 20^\circ )} \right] + \dfrac{{\sqrt 3 }}{8}\sin 20^\circ$
Now, $\sin ( - \theta ) = - \sin \theta$
$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\sin 60^\circ - \sin 20^\circ } \right) + \dfrac{{\sqrt 3 }}{8}\sin 20^\circ$
$\Rightarrow \dfrac{{\sqrt 3 }}{8}\sin 60^\circ - \dfrac{{\sqrt 3 }}{8}\sin 20^\circ + \dfrac{{\sqrt 3 }}{8}\sin 20^\circ$
Subtracting the terms and substituting the value of $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$ we get,
$\Rightarrow \dfrac{{\sqrt 3 }}{8} \times \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \dfrac{3}{{16}}$
=R.H.S.
L.H.S. = R.H.S.
$\therefore$Hence, $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{{16}}$(Proved).

Note: Sin Cos formulas are based on sides of the right-angled triangle. Sin and Cos are basic trigonometric functions along with tan function, in trigonometry. Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse.

${\text{sin}} \theta = \dfrac{{{\text{Opposite side}}}}{{{\text{Hypotenuse}}}}$
${\text{cos}} \theta = \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}$
In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. The most widely used trigonometric functions are the sine, the cosine, and the tangent.