RD Chapter 4 Inverse Trigonometric Functions Ex 4.10 Solutions
Question - 1 : - Evaluate:
(i) Cot (sin-1 (3/4) + sec-1 (4/3))
(ii) Sin (tan-1 x + tan-1 1/x) for x <0
(iii) Sin (tan-1 x + tan-1 1/x) for x> 0
(iv) Cot (tan-1 a + cot-1 a)
(v) Cos (sec-1 x + cosec-1 x), |x| ≥ 1
Answer - 1 : -
(i) Given Cot (sin-1 (3/4)+ sec-1 (4/3))
(ii) Given Sin (tan-1 x+ tan-1 1/x) for x < 0
(iii) Given Sin (tan-1 x+ tan-1 1/x) for x > 0
(iv) Given Cot (tan-1 a+ cot-1 a)
(v) Given Cos (sec-1 x+ cosec-1 x), |x| ≥ 1
= 0
If cos-1 x + cos-1 y = π/4, find thevalue of sin-1 x + sin-1 y.
Answer - 2 : -
Given cos-1 x+ cos-1 y = π/4
Question - 3 : - If sin-1 x + sin-1 y= π/3 and cos-1 x – cos-1 y = π/6, find thevalues of x and y.
Answer - 3 : -
Given sin-1 x+ sin-1 y = π/3 ……. Equation (i)
And cos-1 x– cos-1 y = π/6 ……… Equation (ii)
Question - 4 : - If cot (cos-1 3/5 + sin-1 x) = 0, findthe value of x.
Answer - 4 : -
Given cot (cos-1 3/5+ sin-1 x) = 0
On rearranging we get,
(cos-1 3/5+ sin-1 x) = cot-1 (0)
(Cos-1 3/5+ sin-1 x) = π/2
We know that cos-1 x+ sin-1 x = π/2
Then sin-1 x= π/2 – cos-1 x
Substituting the abovein (cos-1 3/5 + sin-1 x) = π/2 we get,
(Cos-1 3/5+ π/2 – cos-1 x) = π/2
Now on rearranging weget,
(Cos-1 3/5– cos-1 x) = π/2 – π/2
(Cos-1 3/5– cos-1 x) = 0
Therefore Cos-1 3/5= cos-1 x
On comparing the aboveequation we get,
x = 3/5
Question - 5 : - If (sin-1 x)2 +(cos-1 x)2 = 17 π2/36, find x.
Answer - 5 : -
Given (sin-1 x)2 +(cos-1 x)2 = 17 π2/36
We know that cos-1 x+ sin-1 x = π/2
Then cos-1 x= π/2 – sin-1 x
Substituting this in (sin-1 x)2 +(cos-1 x)2 = 17 π2/36 we get
(sin-1 x)2 +(π/2 – sin-1 x)2 = 17 π2/36
Let y = sin-1 x
y2 +((π/2) – y)2 = 17 π2/36
y2 + π2/4– y2 – 2y ((π/2) – y) = 17 π2/36
π2/4 – πy +2 y2 = 17 π2/36
On rearranging andsimplifying, we get
2y2 –πy + 2/9 π2 = 0
18y2 –9 πy + 2 π2 = 0
18y2 –12 πy + 3 πy + 2 π2 = 0
6y (3y – 2π) + π (3y –2π) = 0
Now, (3y – 2π) = 0 and(6y + π) = 0
Therefore y = 2π/3 andy = – π/6
Now substituting y = –π/6 in y = sin-1 x we get
sin-1 x= – π/6
x = sin (- π/6)
x = -1/2
Now substituting y =-2π/3 in y = sin-1 x we get
x = sin (2π/3)
x = √3/2
Now substituting x =√3/2 in (sin-1 x)2 + (cos-1 x)2 =17 π2/36 we get,
= π/3 + π/6
= π/2 which is notequal to 17 π2/36
So we have to neglectthis root.
Now substituting x =-1/2 in (sin-1 x)2 + (cos-1 x)2 =17 π2/36 we get,
= π2/36 + 4π2/9
= 17 π2/36
Hence x = -1/2.
Question - 6 : - Solve:
Answer - 6 : -
sin-1(1/5)+ [ Π/2 - sin-1x ] = sin-11
sin-1(1/5)+ Π/2 - sin-1x = Π/2
sin-1(1/5)- sin-1x = 0
x = 1/5
Question - 7 : - Solve:
Answer - 7 : -
Π/2 - cos-1x= Π/6 + cos-1x
Π/3 = 2cos-1x
cos-1x= Π/6
x = √3/2
Question - 8 : - Solve:
4 sin-1x = Π - cos-1x
Answer - 8 : -
4sin-1x+cos-1x=Π
3sin-1x+sin-1x+cos-1x=Π
3sin-1x=Π/2 [sin-1x+cos-1x=Π/2]
sin-1x=Π/6
x = sinΠ/6=0.5
Question - 9 : - Solve:
Answer - 9 : -
tan-1x+cot-1x=Π/2so the above equation reduces to
cot-1x=2Π/3-Π/2 =Π/6
x= cotΠ/6 =√3
Question - 10 : - Solve:
5 tan-1x + 3 cot-1x = 2Π
Answer - 10 : -
2tan-1x+3(Π/2)=2Π
2tan-1x=2Π-3Π/2=Π/3
tan-1x=Π/6
x=tanΠ/6=1/√3