第12类NCERT解决方案–数学第I部分–第2章反三角函数–练习2.1

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问题1.罪^-1 (-1/2)

解决方案：

Let sin^-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2,π/2] and sin(-π/6)=-1/2.

Therefore, principal value of sin^-1(-1/2)=-π/6.

问题2. cos ^-1 (√3/ 2)

解决方案：

Let cos^-1(√3/2) = y then, cos y = √3/2

Range of principal value for cos^-1 is [0, π] and cos(π/6) = √3/2

Therefore, principal value of cos^-1(√3/2) = π/6.

问题3. cosec ^-1 (2)

解决方案：

Let cosec^-1(2) = y then, cosec y = 2

Range of principal value for cosec^-1 is [-π/2, π/2] -{0}  and cosec(π/6) = 2

Therefore, principal value of cosec^-1(2) = π/6.

问题4：棕褐色^-1 (-√3)

解决方案：

Let tan^-1(-√3) = y then, tan y = -√3

Range of principal value for tan^-1 is (-π/2, π/2) and tan(-π/3) = -√3

Therefore, principal value of tan^-1(-√3) = -π/3.

问题5. cos ^-1 (-1/2)

解决方案：

Let cos^-1(-1/2) = y then, cos y = -1/2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 2π/3.

问题6. tan-1(-1)

解决方案：

Let tan^-1(-1) = y then, tan y = -1

Range of principal value for tan^-1 is (-π/2, π/2)  and tan(-π/4) = -1

Therefore, principal value of tan^-1(-1) = -π/4.

问题7.秒^-1 (2 /√3)

解决方案：

Let sec^-1(2/√3) = y then, sec y = 2/√3

Range of principal value for sec^-1 is [0, π] – {π/2} and sec(π/6) = 2/√3

Therefore, principal value of sec^-1(2/√3) = π/6.

问题8. cot ^-1 (√3)

解决方案：

Let cot^-1(√3) = y then, cot y = √3

Range of principal value for cot^-1 is (0, π) and cot(π/6) = √3

Therefore, principal value of cot^-1(√3) = π/6.

问题9. cos ^-1 (-1 /√2)

解决方案：

Let cos^-1(-1/√2) = y then, cos y = -1/√2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 3π/4.

问题10. cosec ^-1 (-√2)

解决方案：

Let cosec^-1(-√2) = y then, cosec y = -√2

Range of principal value for cosec^-1 is [-π/2, π/2] -{0} and cosec(-π/4) = -√2

Therefore, principal value of cosec^-1(-√2) = -π/4.

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问题11.tan ^-1 (1)+ cos ^-1 (-1/2)+ sin ^-1 (-1/2)

解决方案：

For solving this question we will use principal values of sin^-1, cos^-1 & tan^-1

Let sin^-1(-1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2, π/2] and sin(-π/6) = -1/2.

Therefore, principal value of sin^-1(-1/2) = -π/6.

Let cos^-1(-1/2) = x then, cos x = -1/2

Range of principal value for cos^-1 is [0, π] and cos(2π/3) = -1/2

Therefore, principal value of cos^-1(-1/2) = 2π/3.

Let tan^-1(1) = z then, tan z = -1

Range of principal value for tan^-1 is (-π/2, π/2)  and tan(π/4) = 1

Therefore, principal value of tan^-1(1) = π/4.

Now, tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2) = π/4 + 2π/3 – π/6 

Adding them we will get, 

= (3π + 8π – 2π)/12 

= 9π/12

= 3π/4 

问题12.cos ^-1 (1/2)+ 2 sin ^-1 (1/2)

解决方案：

For solving this question we will use principal values of sin^-1 & cos^-1

Let sin^-1(1/2) = y then, sin y = -1/2

Range of principal value for sin^-1 is [-π/2, π/2] and sin(π/6) = 1/2.

Therefore, principal value of sin^-1(1/2) = π/6.

Let cos^-1(1/2) = x then, cos x = 1/2

Range of principal value for cos^-1 is [0, π] and cos(π/3) = 1/2

Therefore, principal value of cos^-1(1/2) = π/3.

Now, cos^-1(1/2) + 2 sin^-1(1/2) = π/3 + 2π/6 

Adding them we will get,

= (2π + 2π)/6

= 4π/6

= 2π/3 

问题13：如果sin ^–1 x = y，则

(A)0≤y≤π(B)-π/ 2≤y≤π/ 2(C)0

解决方案：

We know that the principal range for sin-1 is [-π / 2, π / 2]

Hence, if sin-1 x = y, y € [-π / 2, π / 2]

Therefore, -π / 2 ≤y ≤ π / 2.

Hence, option (B) is correct.

问题14. tan ^–1 (√3)–秒^-1 (-2)等于

( A)π(B)-π/ 3(C)π/ 3(D)2π/ 3

解决方案：

For solving this question we will use principal values of sec^-1 & tan^-1

Let tan^-1(√3) = y then, tan y = √3

Range of principal value for tan^-1 is (-π/2, π/2) and tan(π/3) = √3

Therefore, principal value of tan^-1(√3) = π/3.

Let sec^-1(-2) = y then, sec y = -2

Range of principal value for sec^-1 is [0, π] – {π/2} and sec(2π/3) = – 2

Therefore, principal value of sec^-1(-2) = 2π/3.

Now, tan^–1 (√3) – sec ^-1(-2) 

= π/3 – 2π/3

= -π/3

Hence, option (B) is correct.