If A=0−tan⁡α2tan⁡α20 and I is the identity matrix of order 2, then (I – A)cos⁡α−sin⁡αsin⁡αcos⁡α is equal to

Here, A=0−tt0, where t=tan⁡α2Now, cos⁡α=1−tan2⁡α21+tan2⁡α2=1−t21+t2and sin⁡α=2tan⁡α21+tan2⁡α2=2t1+t2=(I−A)cos⁡α−sin⁡αsin⁡αcos⁡α=1001−0−t+t01−t21+t2−2t1+t22t1+t21−t21+t2=1t−t11−t21+t2−2t1+t22t1+t21−t21+t2=1−t2+2t21+t2−2t+t1−t21+t2−t1−t2+2t1+t22t2+1−t21+t2=1+t21+t2−2t+t−t31+t2−t+t3+2t1+t22t2+1−t21+t2=1+t21+t2−t1+t21+t2t1+t21+t21+t21+t2=1−tt1Also, I+A=1 00 1+0−tt0=0+1−t+0t+00+1=1−tt1=(I−A)cos⁡α−sin⁡αsin⁡αcos⁡α

If A=0−tan⁡α2tan⁡α20 and I is the identity matrix of order 2, then (I – A)cos⁡α−sin⁡αsin⁡αcos⁡α is equal to

Here, A=0−tt0, where t=tan⁡α2Now, cos⁡α=1−tan2⁡α21+tan2⁡α2=1−t21+t2and sin⁡α=2tan⁡α21+tan2⁡α2=2t1+t2=(I−A)cos⁡α−sin⁡αsin⁡αcos⁡α=1001−0−t+t01−t21+t2−2t1+t22t1+t21−t21+t2=1t−t11−t21+t2−2t1+t22t1+t21−t21+t2=1−t2+2t21+t2−2t+t1−t21+t2−t1−t2+2t1+t22t2+1−t21+t2=1+t21+t2−2t+t−t31+t2−t+t3+2t1+t22t2+1−t21+t2=1+t21+t2−t1+t21+t2t1+t21+t21+t21+t2=1−tt1Also, I+A=1 00 1+0−tt0=0+1−t+0t+00+1=1−tt1=(I−A)cos⁡α−sin⁡αsin⁡αcos⁡α

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $A = [\begin{array}{cc} 0 & - \tan \frac{α}{2} \\ \tan \frac{α}{2} & 0 \end{array}]$ and $I$ is the identity matrix of order 2, then (I – A) $[\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$I - A$

$A - I$

$I + A$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Here, $A = [\begin{array}{cc} 0 & - t \\ t & 0 \end{array}]$ , where $t = \tan (\frac{α}{2})$

Now, $\cos α = \frac{1 - \tan^{2} (\frac{α}{2})}{1 + \tan^{2} (\frac{α}{2})} = \frac{1 - t^{2}}{1 + t^{2}}$

and $s in α = \frac{2 \tan (\frac{α}{2})}{1 + \tan^{2} (\frac{α}{2})} = \frac{2 t}{1 + t^{2}}$

$\begin{array}{l} = (I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \\ = ([\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] - [\begin{array}{cc} 0 & - t \\ + t & 0 \end{array}]) [\begin{array}{cc} \frac{1 - t^{2}}{1 + t^{2}} & \frac{- 2 t}{1 + t^{2}} \\ \frac{2 t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{array}] \\ = [\begin{array}{cc} 1 & t \\ - t & 1 \end{array}] [\begin{array}{cc} \frac{1 - t^{2}}{1 + t^{2}} & \frac{- 2 t}{1 + t^{2}} \\ \frac{2 t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{array}] \end{array}$

$\begin{array}{l} = [\begin{array}{cc} \frac{1 - t^{2} + 2 t^{2}}{1 + t^{2}} & \frac{- 2 t + t (1 - t^{2})}{1 + t^{2}} \\ \frac{- t (1 - t^{2}) + 2 t}{1 + t^{2}} & \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}} \end{array}] \\ = [\begin{array}{cc} \frac{1 + t^{2}}{1 + t^{2}} & \frac{- 2 t + t - t^{3}}{1 + t^{2}} \\ \frac{- t + t^{3} + 2 t}{1 + t^{2}} & \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}} \end{array}] \end{array}$

$= [\begin{array}{cc} \frac{1 + t^{2}}{1 + t^{2}} & \frac{- t (1 + t^{2})}{1 + t^{2}} \\ \frac{t (1 + t^{2})}{1 + t^{2}} & \frac{1 + t^{2}}{1 + t^{2}} \end{array}] = [\begin{array}{cc} 1 & - t \\ t & 1 \end{array}]$

Also, $I + A = [\begin{array}{ll} 1 0 \\ 0 1 \end{array}] + [\begin{array}{cc} 0 & - t \\ t & 0 \end{array}]$

$\begin{array}{l} = [\begin{array}{cc} 0 + 1 & - t + 0 \\ t + 0 & 0 + 1 \end{array}] \\ = [\begin{array}{cc} 1 & - t \\ t & 1 \end{array}] = (I - A) [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}] \end{array}$