第 12 课 NCERT 解决方案 - 数学第一部分 - 第 2 章反三角函数 - 练习 2.2 |设置 1

证明以下

问题1. 3sin ^-1 x = sin ^-1 (3x – 4x ³ ), x∈[-1/2, 1/2]

解决方案：

Let us take x = sinθ, so θ = sin^-1x

Substitute the value of x in the equation present on R.H.S. 

The equation becomes sin^-1(3sinθ – 3sin³θ) 

We know, sin3θ = 3sinθ – 4sin³θ

So , sin^-1(3sinθ – 3sin³θ) = sin^-1(sin3θ) 

By the property of inverse trigonometry we know, sin(sin^-1(θ)) = θ   

So, sin^-1(sin3θ) = 3θ 

And we know θ = sin^-1x   

So, 3θ = 3sin^-1x = L.H.S

问题2. 3cos ^-1 x = cos ^-1 (4x ³ – 3x), x∈[-1/2, 1]

解决方案：

Let us take x = cosθ, so θ = cos^-1x

Substitute value of x in the equation present on R.H.S. 

The equation becomes cos^-1(4cos³θ – 3cosθ) 

We know, cos3θ = 4cos³θ – 3cosθ  

So, cos^-1(4cos³θ – 3cosθ) = cos^-1(cos3θ) 

By the property of inverse trigonometry we know, cos(cos^-1(θ)) = θ    

So, cos^-1(cos3θ) = 3θ 

And we know θ = cos^-1x     

So, 3θ = 3cos^-1x = L.H.S

问题 3。 $tan^{-1}\frac{2}{11} + tan^{-1}\frac{7}{24} = tan^{-1}\frac{1}{2}$

解决方案：

We know, $tan^{-1}x + tan^{-1}y = tan^{-1}\frac{x+y}{1-xy}$

Now put x = 2/11 and y = 7/24

So, $tan^{-1}\frac{2}{11} + tan^{-1}\frac{7}{24} = tan^{-1}\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}.\frac{7}{24}}$

$=tan^{-1}\frac{\frac{48}{264}+\frac{77}{264}}{1-\frac{14}{264}}$

$=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}$

$=tan^{-1}\frac{\frac{48+77}{264}}{\frac{264-14}{264}}$

$=tan^{-1}\frac{125}{250}$

$=tan^{-1}\frac{1}{2}$

= R.H.S

问题 4。 $2tan^{-1}\frac{1}{2} + tan^{-1}\frac{1}{7} = tan^{-1}\frac{31}{17}$

解决方案：

We have to first write 2tan^-1x in terms of  tan^-1x

We know that 2tan^-1x = $tan^{-1}\frac{2x}{1-x^2}$ 

Put x = 1/2 in the above formula

So,  $2tan^{-1}\frac{1}{2} = tan^{-1}\frac{2.\frac{1}{2}}{1-(\frac{1}{2})^2}$

$= tan^{-1}\frac{1}{1-(\frac{1}{4})^2}$

$= tan^{-1}\frac{1}{\frac{4-1}{4}}$

$= tan^{-1}\frac{1}{\frac{3}{4}}$

$= tan^{-1}\frac{4}{3}$

Now we can replace $2tan^{-1}\frac{1}{2}$with $tan^{-1}\frac{4}{3}$

So equation in L.H.S become $tan^{-1}\frac{4}{3} + tan^{-1}\frac{1}{7}$

We know , $tan^{-1}x + tan^{-1}y = tan^{-1}\frac{x+y}{1-xy}$ 

So,  $tan^{-1}\frac{4}{3} + tan^{-1}\frac{1}{7} = tan^{-1}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}$

$=tan^{-1}\frac{\frac{28}{21}+\frac{3}{21}}{1-\frac{4}{21}}$

$=tan^{-1}\frac{\frac{28+3}{21}}{\frac{21-4}{21}}$

$=tan^{-1}\frac{\frac{31}{21}}{\frac{17}{21}}$

$=tan^{-1}\frac{31}{17}$ 

= R.H.S

用最简单的形式写出以下函数：

问题 5。 $tan^{-1}\frac{\sqrt{1+x^2}-1}{x} , x ≠ 0$

解决方案：

Let us assume that x = tanθ, so θ = tan^-1x 

Substitute the value of x in question. 

So equation becomes $tan^{-1}\frac{\sqrt{1+tan^2θ}-1}{tanθ}$

We know that, 1 + tan²θ = sec²θ 

Replacing 1 + tan²θ with sec²θ in the equation $tan^{-1}\frac{\sqrt{1+tan^2θ}-1}{tanθ}$

So equation becomes, $tan^{-1}\frac{\sqrt{sec^2θ}-1}{tanθ}$

$= tan^{-1}\frac{secθ-1}{tanθ}$

We know, tanθ = sinθ/cosθ and sec = 1/cosθ 

Replacing value of tanθ and secθ in $tan^{-1}\frac{secθ-1}{tanθ}$

$= tan^{-1}\frac{\frac{1}{cosθ}-1}{\frac{sinθ}{cosθ}}$

$= tan^{-1}\frac{\frac{1-cosθ}{cosθ}}{\frac{sinθ}{cosθ}}$

$= tan^{-1}\frac{1-cosθ}{sinθ}$

We know, 1 – cosθ = 2sin²θ/2​ and sinθ = 2sinθ/2cosθ/2  

So the equations after replacing above value becomes $tan^{-1}\frac{2sin^2\frac{θ}{2}}{2sin\frac{θ}{2}cos\frac{θ}{2}}$

$= tan^{-1}\frac{sin\frac{θ}{2}}{cos\frac{θ}{2}}$

We know $\frac{\frac{sinθ}{2}}{\frac{cosθ}{2}} = tan\frac{θ}{2}$ 

$= tan^{-1}tan\frac{θ}{2}$

= θ/2           [tan^-1(tanθ) = θ]

= 1/2 tan^-1x            [θ = tan^-1x]   

问题 6。 $tan^{-1}\frac{1}{\sqrt{x^2-1}}$ , |x| > 1

解决方案：

Let us assume that x = cosecθ, so θ = cosec ^-1x 

Substitute the value of x in question with $cosecθ$

$=tan^{-1}\frac{1}{\sqrt{cosec^2θ-1}}$

We know that, 1 + cot²θ = cosec²θ, so cosec²θ = 1 – cot²θ  

$=tan^{-1}\frac{1}{\sqrt{cot^2θ}}$

$=tan^{-1}\frac{1}{cotθ}$

= tan^-1(tanθ)          [1/cotθ = tanθ]  

= θ           [tan^-1(tanθ) = θ] 

= cosec⁻¹x          [θ = cosec⁻¹x]

= π/2 ​- sec⁻¹x         [cosec⁻¹x + sec⁻¹x = π/2​]

问题 7。 $tan^{-1}(\sqrt{\frac{1-cosx}{1+cosx}}), 0<x<\pi$

解决方案：

We know, 1 – cosx = 2sin²x/2 and 1 + cosx = 2cos²x/2   

 Substituting above formula in question

$=tan^{-1}(\sqrt{\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}})$

$=tan^{-1}(\frac{sin\frac{x}{2}}{cos\frac{x}{2}})$

= tan^-1(tanx/2)         $[\frac{sin\theta}{cos\theta} = tan\theta]$

 = x/2         [tan^-1(tanθ) = θ] 

问题 8。 $tan^{-1}(\frac{cosx-sinx}{cosx+sinx}), \frac{-\pi}{4}<x<\frac{3\pi}{4}$

解决方案：

Divide numerator and denominator by $cosx$

$= tan^{-1}(\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}})$

$= tan^{-1}(\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}})$

$= tan^{-1}(\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}})$

We know, $\frac{sinx}{cosx} = tanx$

$= tan^{-1}(\frac{1-tanx}{1+tanx})$

This can also be written as $tan^{-1}(\frac{1-tanx}{1+1.tanx})$  – (1)

We know  $tan^{-1}x-tan^{-1}y = tan^{-1}\frac{x-y}{1+x.y}$      – (2)

On comparing equation (1) and (2) we can say that x = 1 and y = tan^-1x

So we can say that $tan^{-1}(\frac{1-tanx}{1+1.tanx}) = tan^{-1}1 - tan^{-1}x$

= π/4​ – tan⁻¹x          [tan⁻¹1 = π/4​]

问题 9。 $tan^{-1}\frac{x}{\sqrt{a^2-x^2}} , |x|<a$

解决方案：

Let us assume that x = asinθ, so θ = sin ^-1x/a

Substitute the value of x in question.

$= tan^{-1}\frac{asin\theta}{\sqrt{a^2-(asin\theta)^2}}$

$= tan^{-1}\frac{asin\theta}{\sqrt{a^2-a^2sin^2\theta}}$

Taking a² common from denominator

$= tan^{-1}\frac{asin\theta}{\sqrt{a^2(1-sin^2\theta)}}$

We know that, sin²θ + cos²θ = 1, so 1 – sin²θ = cos²θ 

$= tan^{-1}\frac{asin\theta}{\sqrt{a^2cos^2\theta}}$

$= tan^{-1}\frac{asin\theta}{acos\theta}$

$= tan^{-1}\frac{sin\theta}{cos\theta}$

= tan^-1(tanθ)          [sinθ/cosθ = tanθ]

 = θ        

= sin^-1x/a 

问题 10。 $tan^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2}),a>0;\frac{-a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}$

解决方案：

Let us assume that x = atanθ, so θ = tan ^-1x/a

Substitute the value of x in question

 $= tan^{-1}(\frac{3a^2(atan\theta)-(atan\theta)^3}{a^3-3a(atan\theta)^2})$

$= tan^{-1}(\frac{3a^3tan\theta-a^3tan^3\theta}{a^3-3a^3tan^2\theta})$

Taking a³common from numerator and denominator

$= tan^{-1}(\frac{a^3(3tan\theta-tan^3\theta)}{a^3(1-3tan^2\theta)})$

$= tan^{-1}(\frac{3tan\theta-tan^3\theta}{1-3tan^2\theta})$

We know $tan3\theta = \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}$

So, $tan^{-1}(\frac{3tan\theta-tan^3\theta}{1-3tan^2\theta})= tan^{-1}(tan3\theta)$

= 3θ           [ tan^-1(tanθ) = θ]

= 3tan ^-1x/a

第 2 章反三角函数 – 练习 2.2 |设置 2