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Class 12 Question Paper Alternative English Vibyor AHSEC 2017

Alternative English  Year - 2017 Full Marks : 100 
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Class 12 Maths NCERT solutions Inverse Trigonometric Functions


  Inverse Trigonometric Functions.    
Chapter 2

                                                       Exercise 2.1

 1. Find the principal value of sin-1 (-\( \frac{1}{2} \))
Sol. Let sin-1(-\( \frac{1}{2} \))=y. Then sin y =-\(\frac{1}{2}\)
Since the range of the principal value branch of sin-1 is
(-\(\frac{\pi}{2}\,\frac{\pi}{2}\)) and  sin(-\(\frac{\pi}{6}\))=-\(\frac{1}{2}\).Therefore, the principal value of sin-1 (-\( \frac{1}{2} \))is -\(\frac{\pi}{6}\)


2. Find the principal value of cos-1 (\( \frac{\sqrt3}{2} \))
Sol.  Let cos-1(\( \frac{\sqrt3}{2} \))=y. Then cos y =\(\frac{\sqrt3}{2}\)
Since the range of the principal value branch of cos-1 is (0,\(\pi\))
 and  cos(\(\frac{\pi}{6}\))=\(\frac{\sqrt3}{2}\).
Therefore, the principal value of cos-1 (\( \frac{\sqrt3}{2} \))is \(\frac{\pi}{6}\).

3. Find the principal value of cosec-1 (2).
Sol.  Let cosec-1 (2)=y. Then cosec y =2
Since the range of the principal value branch of cosec-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\))-{0}
 and  cosec(\(\frac{\pi}{6}\))=2.
Therefore, the principal value of cosec-1 (2 ) is \(\frac{\pi}{6}\).




4. Find the principal value of tan-1 (\(-\sqrt3)\).
Sol.  Let tan-1 (\(-\sqrt3)\)=y. Then tan y =(\(-\sqrt3)\)
Since the range of the principal value branch of tan-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\))
 and - tan(\(\frac{\pi}{3}\))=(\(-\sqrt3)\).
tan(\(-\frac{\pi}{3}\))=(\(-\sqrt3)\).
Therefore, the principal value of tan-1 (\(-\sqrt3)\ ) is (-\frac{\pi}{3})\).


5. Find the principal value of cos-1 (-\( \frac{1}{2} \))
Sol.  Let cos-1(-\( \frac{1}{2} \))=y. Then cos y =-\(\frac{1}{2}\)
Since the range of the principal value branch of cos-1 is (0,\(\pi\))
and - cos(\(\frac{\pi}{3}\))=-\(\frac{1}{2}\)
which implies - cos(\(\frac{\pi}{3}\))=cos(\(\pi\)-\(\frac{\pi}{3}\))=cos(\(\frac{2\pi}{3}\))

Therefore, the principal value of cos-1 (-\( \frac{1}{2} \))is \(\frac{2\pi}{3}\).


6. Find the principal value of tan-1 (-1)
Sol.  Let tan-1 (-1)=y. Then tan y =-1
Since the range of the principal value branch of tan-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\))
 and  -tan(\(\frac{\pi}{4}\))=-1.
tan(-\(\frac{\pi}{4}\))=-1

Therefore, the principal value of tan-1 (-1 ) is -\(\frac{\pi}{4}\).



7. Find the principal value of sec-1 (\( \frac{2}{\sqrt3} \))
Sol.  Let sec-1 (\( \frac{2}{\sqrt3} \))=y. Then sec y =\( \frac{2}{\sqrt3} \)
Since the range of the principal value branch of sec-1 is (0,\(\pi\))-{ \(\frac{\pi}{2}\)}
and  sec(\(\frac{\pi}{6}\))=\( \frac{2}{\sqrt3} \)
Therefore, the principal value of sec1 (\( \frac{2}{\sqrt3} \)) is \(\frac{\pi}{6}\).


8. Find the principal value of cot-1 (\(\sqrt3)\).

Sol.  Let cot-1 (\(\sqrt3)\)=y. Then tan y =(\(\sqrt3)\)
Since the range of the principal value branch of cot-1 is (0,\(\pi\))
 and  cot(\(\frac{\pi}{6}\))=(\(\sqrt3)\).
Therefore, the principal value of cot-1 (\(\sqrt3)\) ) is  \(\frac{\pi}{6}\).


9. Find the principal value of cos-1 (-\( \frac{1}{\sqrt2} \))
Sol. Let cos-1(-\( \frac{1}{\sqrt2} \))=y. Then cos y =-\(\frac{1}{\sqrt2}\)
Since the range of the principal value branch of cos-1 is (0,\(\pi\))
 and  -cos(\(\frac{\pi}{4}\))=-\(\frac{1}{\sqrt2}\).
-cos(\(\frac{\pi}{4}\))=cos(\(\pi\)-\(\frac{\pi}{4}\))=cos(\(\frac{3\pi}{4}\))
Therefore, the principal value of cos-1 (-\( \frac{1}{\sqrt2} \))is \(\frac{3\pi}{4}\).



10. Find the principal value of cosec-1 (\(-\sqrt2)\).
Sol.  Let tcosec-1 (\(-\sqrt2)\). Then cosec y =(\(-\sqrt2)\)
Since the range of the principal value branch of cosec-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\))-{0}
 and  -cosec(\(\frac{\pi}{4}\))=(\(-\sqrt2)\).
 cosec(-\(\frac{\pi}{4}\))=(\(-\sqrt2)\).
Therefore, the principal value of cosec-1 (\(-\sqrt2)\)  is \(-\frac{\pi}{4}\).



Find the values of the following :

11. tan-1(1) + cos-1(\(-\frac{1}{2})\)+ sin-1 (-\( \frac{1}{2} \))

 Sol.  We have to find the principal value of each function as before and add them.
  Let tan-1 (1)=y.Thus, tan y =1=tan \(\frac{\pi}{4}\).
Hence the principal value of tan-1 (1)=\(\frac{\pi}{4}\).
Let  cos-1(\(-\frac{1}{2})\)=y. Thus, cos y =\(-\frac{1}{2}\).
-cos(\(\frac{\pi}{3})\)= \(-\frac{1}{2}\)
But the range of the principal value branch of cos-1 is (0,\(\pi\)).
Hence -cos(\(\frac{\pi}{3})\)= cos(\(\pi\)-\(\frac{\pi}{3})\)=\(-\frac{1}{2}\)
cos(\(\frac{2\pi}{3})\)=\(-\frac{1}{2}\).
Therefore, the principal value of cos-1(\(-\frac{1}{2})\) is \(\frac{2\pi}{3}\)

Let sin-1 (-\( \frac{1}{2} \)) = y. Thus,  sin y =(-\( \frac{1}{2} \)).
-sin(\(\frac{\pi}{6}\))=(-\( \frac{1}{2} \)).
sin(\(-\frac{\pi}{6}\))=(-\( \frac{1}{2} \)).
Therefore, the principal value of sin-1 (-\( \frac{1}{2} \)) = \(-\frac{\pi}{6}\)
Therefore, 
tan-1(1) + cos-1(\(-\frac{1}{2})\)+ sin-1 (-\( \frac{1}{2} \))
= \(\frac{\pi}{4}\)  +\(\frac{2\pi}{3}\)  -\(\frac{\pi}{6}\)
 =\(\frac{3\pi+8\pi-2\pi}{12}\)
=\(\frac{9\pi}{12}\)
=\(\frac{3\pi}{4}\)



12. cos-1(\(\frac{1}{2})\)+ 2sin-1 (\( \frac{1}{2} \))

Sol.  Let  cos-1(\(\frac{1}{2})\)=y. Thus, cos y =\(\frac{1}{2}\).
cos(\(\frac{\pi}{3})\)= \(\frac{1}{2}\)
The range of the principal value branch of cos-1 is (0,\(\pi\)).
Hence,  cos(\(\frac{\pi}{3})\)=\(\frac{1}{2}\)

Therefore, the principal value of cos-1(\(-\frac{1}{2})\) is \(\frac{\pi}{3}\)

Let sin-1 (\( \frac{1}{2} \)) = y. Thus,  sin y =(\( \frac{1}{2} \)).
sin(\(\frac{\pi}{6}\))=(\( \frac{1}{2} \)).
sin(\(-\frac{\pi}{6}\))=(\( \frac{1}{2} \)).
Therefore, the principal value of sin-1 (\( \frac{1}{2} \)) = \(\frac{\pi}{6}\)
Therefore, 
 cos-1(\(\frac{1}{2})\)+ 2sin-1 (\( \frac{1}{2} \))
= \(\frac{\pi}{3}\)  +\(\frac{2\pi}{6}\)
= \(\frac{\pi}{3}\)  +\(\frac{\pi}{3}\)
=\(\frac{2\pi}{3}\)



13. If sin-1 x=y, then
(a) 0≤ y ≤π         (b)-\(\frac{\pi}{2}\)≤ y ≤\(\frac{\pi}{2}\)        (c)0< y <π        (d)-\(\frac{\pi}{2}\)< y <\(\frac{\pi}{2}\)
Sol.   We have been given that sin-1 x=y.Since, the range of principal value branch of  sin-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\)). The correct answer is  (b) -\(\frac{\pi}{2}\)≤ y ≤\(\frac{\pi}{2}\)  .

13.  tan-1 (\(-\sqrt3)\) - sec-1 (-2) is equal to

 (a) π         (b)-\(\frac{\pi}{3}\)        (c)\(\frac{\pi}{3}\)        (d)\(\frac{2\pi}{3}\)

 Sol.  Let tan-1 (\(-\sqrt3)\)=y. Then tan y =(\(-\sqrt3)\)
Since the range of the principal value branch of tan-1 is (-\(\frac{\pi}{2}\,\frac{\pi}{2}\))
 and - tan(\(\frac{\pi}{3}\))=(\(-\sqrt3)\).
tan(\(-\frac{\pi}{3}\))=(\(-\sqrt3)\).
∴ The principal value of tan-1 (\(-\sqrt3)\) is \(\frac{\pi}{3}\)     

 Let sec-1 (-2)=y. Then,  sec y =-2
Since the range of the principal value branch of is (0,\(\pi\))-{ \(\frac{\pi}{2}\)}
 and -sec  \(\frac{\pi}{3}\)=-2
⇒ sec(π-  \(\frac{\pi}{3}\))=-2
⇒ sec( \(\frac{2\pi}{3}\))=-2
∴The principal value of sec-1 (-2)= \(\frac{2\pi}{3}\)

Hence,
tan-1 (\(-\sqrt3)\) - sec-1 (-2)
=\(\frac{\pi}{3}\) - \(\frac{2\pi}{3}\)
=-\(\frac{\pi}{3}\)    
The correct answer is (b)

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Subject : General English


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