Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $\left[ {\dfrac{{\left( {2\sin\alpha } \right)}}{{\left( {1 + \cos\alpha + \sin\alpha } \right)}}} \right] = x$ , then what is the value of $\left[ {\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}}} \right]$?
A. $\dfrac{1}{x}$
B. $x$
C. $1 - x$
D. $1 + x$

seo-qna
Last updated date: 19th Jul 2024
Total views: 62.7k
Views today: 1.62k
Answer
VerifiedVerified
62.7k+ views
Hint: Simplify the given equation by using the trigonometric identities of $\sin 2A$, and $2\cos^{2}A$. After that, use the trigonometric identities $2\sin^{2}A$, and $\cos^{2}A + \sin^{2}A = 1$ to rewrite the required expression. In the end, simplify the new expression and get the required answer.

Formula Used:
$\sin2A = 2\sin A \cos A$
$2\cos^{2}A = 1 + \cos2A$
$2\sin^{2}A = 1 - \cos2A$
$\cos^{2}A + \sin^{2}A = 1$

Complete step by step solution:
The given equation is $\left[ {\dfrac{{\left( {2\sin\alpha } \right)}}{{\left( {1 + \cos\alpha + \sin\alpha } \right)}}} \right] = x$.
Let’s simplify the above equation by using the trigonometric identities $\sin2A = 2\sin A \cos A$ and $2\cos^{2}A = 1 + \cos 2A$.
$\dfrac{2\left(\:2\sin \dfrac{\alpha \:}{2}\cos \dfrac{\alpha \:}{2}\:\right)}{2\cos ^2\dfrac{\alpha \:}{2}\:+\:2\sin \dfrac{\alpha \:}{2}\cos \dfrac{\alpha \:}{2}}\:=\:x$
$ \Rightarrow \dfrac{{4\sin\dfrac{\alpha }{2}\cos\dfrac{\alpha }{2}}}{{2\cos\dfrac{\alpha }{2}\left( {\cos\dfrac{\alpha }{2} + 2\sin\dfrac{\alpha }{2}} \right)}} = x$
Cancel out the common factor.
$\dfrac{{2\sin\dfrac{\alpha }{2}}}{{\cos\dfrac{\alpha }{2} + \sin\dfrac{\alpha }{2}}} = x$ $.....\left( 1 \right)$

Now consider the given expression $\left[ {\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}}} \right]$.
Rewrite the expression by using the trigonometric identities $2\sin^{2}A = 1 - \cos2A$, and $\cos^{2}A +\sin^{2}A = 1$.
$\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}} = \dfrac{{\left( {2\sin^{2}\dfrac{\alpha }{2} + 2\sin\dfrac{\alpha }{2}\cos\dfrac{\alpha }{2}} \right)}}{{\left( {\cos^{2}\dfrac{\alpha }{2} + \sin^{2}\dfrac{\alpha }{2} + 2\sin\dfrac{\alpha }{2}\cos\dfrac{\alpha }{2}} \right)}}$
Simplify the right-hand side.
$\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}} = \dfrac{{2\sin\dfrac{\alpha }{2}\left( {\sin\dfrac{\alpha }{2} + \cos\dfrac{\alpha }{2}} \right)}}{{{{\left( {\cos\dfrac{\alpha }{2} + \sin\dfrac{\alpha }{2}} \right)}^2}}}$
Cancel out the common factors.
$\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}} = \dfrac{{2\sin\dfrac{\alpha }{2}}}{{\left( {\cos\dfrac{\alpha }{2} + \sin\dfrac{\alpha }{2}} \right)}}$
Now compare the above equation with the equation $\left( 1 \right)$.
We get,
$\dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}} = \dfrac{{2\sin\dfrac{\alpha }{2}}}{{\left( {\cos\dfrac{\alpha }{2} + \sin\dfrac{\alpha }{2}} \right)}} = x$
$ \Rightarrow \dfrac{{\left( {1 - \cos\alpha + \sin\alpha } \right)}}{{\left( {1 + \sin\alpha } \right)}} = x$

Option ‘C’ is correct

Note: Do not get confused while using the general trigonometric identities for the half-angle or for the multiple of an angle. To write the identities properly, check the proportions of the angles in the general trigonometric identities.