Question

If $sin\theta +cos\theta =a,cos\theta -sin\theta =b$, then $sin\theta (sin\theta -cos\theta )+si{n}^2\theta (si{n}^2\theta -co{s}^2\theta )+si{n}^3\theta (si{n}^3\theta -co{s}^3\theta )+....$ is equal to

• A. $\frac{1-ab}{1+ab}$
• B. $\frac{1-a^2}{3-a^2}$
• C. $\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$
• D. $\frac{1+ab}{1-ab}+\frac{a^2-1}{3-a^2}$

Answer 1

The correct option is C $\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$

$a^2=1+2sin\theta \ast cos\theta$ and $b^2=1-2sin\theta \ast cos\theta$ and $ab=co{s}^2\theta -si{n}^2\theta$
Separating the question into two
$si{n}^2\theta +si{n}^4\theta +si{n}^6\theta .....-(sin\theta \ast cos\theta +si{n}^2\theta \ast co{s}^2\theta +si{n}^3\theta \ast co{s}^3\theta )....$
This is infinite geometric progression and Formula for it is a/(1-r)
Now taking first part of this
$si{n}^2\theta +si{n}^4\theta +si{n}^6\theta ....=\frac{si{n}^2\theta }{1-si{n}^2\theta }=\frac{si{n}^2\theta }{co{s}^2\theta }$
$1-ab=2si{n}^2\theta$ and $1+ab=2co{s}^2\theta$
Therefore $si{n}^2\theta +si{n}^4\theta +si{n}^6\theta ....=\frac{si{n}^2\theta }{co{s}^2\theta }=\frac{1-ab}{1+ab}$
Second part
$sin\theta \ast cos\theta +si{n}^2\theta \ast co{s}^2\theta +si{n}^3\theta \ast co{s}^3\theta +....=\frac{sin\theta \ast cos\theta }{1-sin\theta \ast cos\theta }$
Multiply and divide by 2 => $2sin\theta \ast cos\frac{\theta }{2}-2sin\theta \ast cos\theta =\frac{a^2-1}{2-(a^2-1)}=\frac{a^2-1}{3-a^2}$
Finally combining both the above equ
$si{n}^2\theta +si{n}^4\theta +si{n}^6\theta .....-(sin\theta \ast cos\theta +si{n}^2\theta \ast co{s}^2\theta +si{n}^3\theta \ast co{s}^3\theta )...=\frac{1-ab}{1+ab}-\frac{(a^2-1)}{(3-a^2)}=\frac{1-ab}{1+ab}+\frac{(1-a^2)}{(3-a^2)}$