मात्रात्मक योग्यता प्रश्न 2428
प्रश्न: $(2{{\cos }^{2}}\theta -1)( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } )$ का मान
विकल्प:
A) 4
B) 1
C) 3
D) 2
Show Answer
उत्तर:
सही उत्तर: C
हल:
- $(2{{\cos }^{2}}\theta -1)( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } )$
$=(2{{\cos }^{2}}\theta -1)( \frac{{{(1+\tan \theta )}^{2}}+{{(1-tan\theta )}^{2}}}{1-{{\tan }^{2}}\theta } )$
$=(2{{\cos }^{2}}\theta -1)[ \frac{ \begin{aligned} & (1^{2}+{{\tan }^{2}}\theta +2\tan \theta ) \ & +,(1^{2}+{{\tan }^{2}}\theta -2\tan \theta \ \end{aligned}}{1-\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} ]$
$=(2{{\cos }^{2}}\theta -1)[ \frac{ \begin{aligned} & (1+{{\tan }^{2}}\theta +2\tan \theta ) \ & +,(1+{{\tan }^{2}}\theta -2\tan \theta \ \end{aligned}}{\frac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} ]$
$=(2{{\cos }^{2}}\theta -1)[ \frac{2,(1+{{\tan }^{2}}\theta ).{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta } ]$
$=(2{{\cos }^{2}}\theta -1),\frac{2{{\sec }^{2}}\theta .{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta ,-(1-{{\cos }^{2}}\theta )}$
$[\because 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ]$
$=(2{{\cos }^{2}}\theta -1)\frac{2{{\sec }^{2}}\theta \frac{1}{{{\sec }^{2}}\theta }}{2{{\cos }^{2}}\theta -1}=2$
BETA
