第 10 类 RD Sharma 解决方案 - 第 5 章三角比 - 练习 5.1 |设置 3

问题 22. 如果 sinθ = a/b，根据 a 和 b 求 secθ + tanθ。

解决方案：

Given:

sinθ = a/b

From Pythagoras theorem,

AC² = BC² + AB²

b² = a² + AB²

AB²=$\sqrt{(b^2-a^2)}$  

Now, 

= secθ + tanθ

=$\frac{b}{\sqrt(b^2-a^2)}+\frac{a}{\sqrt(b^2-a^2)}$

=$\frac{b+a}{\sqrt(b^2-a^2)}$

=$\frac{(\sqrt(b+a))^2}{\sqrt(b-a)\sqrt(b+a)}$

=$\frac{\sqrt(b+a)}{\sqrt(b-a)}$

=$\sqrt\frac{b+a}{b-a}$

问题 23. 如果 8tanA = 15，求 sin A - cos A。

解决方案：

Given:

8tanA = 15

tanA = 15/8

From pythagoras theorem,

AC² = BC² + AB²

AC² = 15² + 8²

AC² = 225 + 64 = 289

AC = 17

Now, 

= sin A − cos A

= 15/17 – 8/17

= (15 – 8)/17

= 7/17  

问题 24. 如果 tanθ = 20/21，证明$\frac{1−sinθ+cosθ}{1+sinθ+cosθ}=\frac{3}{7}$ .

解决方案：

Given: tanθ = 20/21

From Pythagoras theorem,

AC² = BC² + AB²

AC² = 20² + 21²

AC² = 400 + 441 = 841

AC = 29

Now, taking LHS

=$\frac{1−sinθ+cosθ}{1+sinθ+cosθ}$

=$\frac{1−\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}$

=$\frac{29-20+21}{29+20+21}$

= 30/70

= 3/7

问题 25. 如果 cosec A = 2，求$\frac{1}{tanA}+\frac{sinA}{1+cosA}$ .

解决方案：

Given:

cosec A = 2

We know 

sin A = 1/cosecA = 1/2

And, sin 30° = 1/2

A = 30°

tan30° = 1/√3 and cos30° = √3/2 

Now,

=$\frac{1}{tanA}+\frac{sinA}{1+cosA}$

=$\frac{1}{\frac{1}{\sqrt3}}+\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$

=$\sqrt3+\frac{\frac{1}{2}}{\frac{2+\sqrt3}{2}}$

=$\sqrt3+\frac{1}{2+\sqrt3}$

=$\frac{2\sqrt3+3+1}{2+\sqrt3}$
=$\frac{2\sqrt3+4}{2+\sqrt3}$  

= 2 

问题 26. 如果∠A和∠B是锐角，使得cos A = cos B，则证明∠A = ∠B。

解决方案：

Let us consider a △ABC 

From the figure,

Given,

cos A = cos B

AC/AB = BC/AB

Multiplying both side by AB

(AC/AB) × AB = (BC/AB) × AB 

AC = BC

In △ABC, AC = BC So we can say that the triangle is an isosceles triangle,

and in an isosceles triangle we know that if two sides of a triangle are equal, 

then the angle opposite to the sides are equal.

Therefore, ∠A =∠B

问题 27. 在与 A 成直角的 Δ ABC 中，如果 tanC = √3，求 sin B cos C + cos B sin C 的值。

解决方案：

In right angled Δ ABC,

Given: tan C = √3 

∴AB = √3 and AC = 1

From pythagoras theorem,

BC² = AB² + AC²

BC² = (√3)² + 1²

BC² = 3 + 1 = 4

BC = 2

Therefore,

sin B cos C + cos B sin C

= (1/2)(√3/2) + (√3/2)(√3/2)

= 1/4 + 3/4

= 4/4

= 1

问题 28. 说明以下是对还是错。证明你的答案。

(i) tan A 的值始终小于 1。

解决方案：

FALSE. The value of tan A is not always less than 1.

Consider the Pythagorean triplet, 13, 12, and 5

where, 13 is the hypotenuse

We know

tan A = Perpendicular/Base

Let Perpendicular = 12 and Base = 5

then, tanA = 12/5 = 2.4 which is greater than 1.

(ii) 对于角度 A 的某个值，sec A = 12/5。

解决方案：

TRUE 

We have sec A = 12/5 for some value of ∠A

secθ = Hypotenuse/Base 

In a right angled triangle, hypotenuse is the greatest side.

So secθ > 1 is valid 

Here, secθ = 12/5  

(iii) cos A 是角 A 的余割线的缩写。

解决方案：

FALSE 

cos A means cosine of ∠A

cos A = Base/Hypotenuse 

However,

cosec A = Hypotenuse/Perpendicular      

(iv) cot A 是 cot 和 A 的乘积。

解决方案：

FALSE

cot A means Cotangent of ∠A

cot A = 1/tanA

Only “cot” doesn’t defines anything.

Hence, cot A is not the product of cot and A.    

(v) 对于某个角度 θ，sinθ = 4/3。

解决方案：

FALSE

sinθ = 4/3 for some value of ∠θ

We have,

sinθ = Perpendicular/Hypotenuse 

In a right angled triangle, hypotenuse is the greatest side.

So sinθ is always less than 1.

Here, sinθ = 4/3 = 1.3 which is greater than 1 

问题 29. 如果 sinθ = 12/13，求$\frac{sin^2θ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}$ .

解决方案：

Given:

sinθ = 12/13

Using Pythagoras theorem,

AC² = BC² + AB²

13² = 12² + AB²

AB² = 169 − 144 = 25

AB = 5

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

=$\frac{(\frac{12}{13})^2-(\frac{5}{13})^2}{2(\frac{12}{13})(\frac{5}{13})}×\frac{1}{(\frac{12}{5})^2}$

=$\frac{\frac{144}{169}-\frac{25}{169}}{\frac{120}{169}}×\frac{25}{144}$

=$\frac{\frac{144-25}{169}}{\frac{120}{169}}×\frac{25}{144}$

=$\frac{119}{120}×\frac{25}{144}$

= 595/3456 

问题 30. 如果 cosθ = 5/13，求$\frac{sinθ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}$ .

解决方案：

Given:

cosθ = 5/13

Using Pythagoras theorem,

AC² = BC² + AB²

13² = BC² + 52

BC² = 169 − 25 = 144

BC = 12

= $\frac{sinθ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}$

=$\frac{(\frac{12}{13})-(\frac{5}{13})^2}{2(\frac{12}{13})(\frac{5}{13})}×\frac{1}{(\frac{12}{5})^2}$

= $\frac{\frac{144}{169}-\frac{25}{169}}{\frac{120}{169}}×\frac{25}{144}$

=$\frac{\frac{144-25}{169}}{\frac{120}{169}}×\frac{25}{144}$

= $\frac{119}{120}×\frac{25}{144}$

= 595/3456 

问题 31. 如果 secA = 5/4，验证$\frac{3sinA−4sin^3A }{4cos^3A−3cosA}=\frac{3tanA−tan^3A}{1−3tan^2A}$ .

解决方案：

Given:

 secA = 5/4

From pythagoras theorem,

AC² = BC² + AB²

5² = BC² + 4²

BC² = 25 − 16 = 9

BC = 3

Now 

=$\frac{3sinA−4sin^3A }{4cos^3A−3cosA}=\frac{3tanA−tan^3A}{1−3tan^2A}$

=$\frac{3(\frac{3}{5})−4(\frac{3}{5})^3 }{4(\frac{4}{5})^3−3(\frac{4}{5})}=\frac{3(\frac{3}{4})−(\frac{3}{4})^3}{1−3(\frac{3}{4})^2}$

=$\frac{(\frac{9}{5})−(\frac{108}{125}) }{(\frac{256}{125})−(\frac{12}{5})}=\frac{(\frac{9}{4})−(\frac{27}{64})}{1−(\frac{27}{16})}$

=$\frac{(\frac{225-108}{125}) }{(\frac{256-300}{125})}=\frac{(\frac{144-27}{64})}{(\frac{16-27}{16})}$

=$\frac{(\frac{117}{125}) }{(\frac{-44}{125})}=\frac{(\frac{117}{64})}{(\frac{-11}{16})}$

= 117/-44 = 117/(11(4))

= 117/-44 = 117/-44

Hence Proved

问题 32. 如果 sinθ = 3/4，证明$\sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1}=\frac{\sqrt 7}{3}$ .

解决方案：

Given: sinθ = 3/4 

From Pythagoras theorem,

AC² = BC² + AB²

4² = AB² + 3²

AB² = 16 – 9 = 7

AB =√7

We have,

$\sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1}=\frac{\sqrt 7}{3}$

 Now squaring both side

=$(\sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1})^2=(\frac{\sqrt 7}{3})^2$

=$\frac{cosec^2θ−cot^2θ}{sec^2θ−1}$ 

= 7/9

We know

1 + cot²θ = cosec²θ

1 + tan²θ = sec²θ

= 1/tan²θ = 7/9 

=$\frac{1}{(\frac{3}{\sqrt7})^2}=\frac{7}{9}$

= 7/9 = 7/9 

Hence Proved

问题 33. 如果 secA = 17/8，验证$\frac{3−4sin^2A}{4cos^2A−3}=\frac{3−tan^2A}{1−3tan^2A}$ .

解决方案：

Given: secA = 17/8

From Pythagoras theorem,

AC² = BC² + AB²

17² = BC² + 8²

BC² = 289 − 64 = 225

BC = 15$\frac{(\frac{-33}{289})}{(\frac{-611}{289})}=\frac{(\frac{-33}{64})}{(\frac{-611}{64})}$

We have 

$\frac{3−4sin^2A}{4cos^2A−3}=\frac{3−tan^2A}{1−3tan^2A}$

Putting the values of sinA, cosA and tanA in the above equation 

=$\frac{3−4\frac{15}{17}^2}{4\frac{8}{17}^2−3}=\frac{3−\frac{15}{8}^2}{1−3\frac{15}{8}^2}$

=$\frac{3−(\frac{900}{289})}{(\frac{256}{289})−3}=\frac{3−(\frac{225}{64})}{1−(\frac{675}{64})}$

=$\frac{(\frac{867-900}{289})}{(\frac{256-867}{289})}=\frac{(\frac{192-225}{64})}{(\frac{64-675}{64})}$

=$\frac{(\frac{-33}{289})}{(\frac{-611}{289})}=\frac{(\frac{-33}{64})}{(\frac{-611}{64})}$

= 33/611 = 33/611

Hence Proved

问题 34. 如果 cotθ = 3/4，证明$\sqrt\frac{secθ−cosecθ}{secθ+cosecθ}=\frac{1}{\sqrt 7}$ .

解决方案：

Given:  cotθ = 3/4

 tanθ = 4/3

Using the pythagoras theorem 

 sinθ = 4/5, cosθ = 3/5

cosecθ = 5/4, secθ = 5/3

Now, taking LHS

=$\sqrt\frac{secθ−cosecθ}{secθ+cosecθ}$

=$\sqrt\frac{\frac{5}{3}−\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}$

=$\sqrt\frac{\frac{5}{12}}{\frac{35}{12}}$

=$\sqrt\frac{5}{35}$  

= 1/√7

问题 35. 如果 3cosθ - 4sinθ = 2cosθ + sinθ，则求 tanθ。

解决方案：

Given: 3cosθ − 4sinθ = 2cosθ + sinθ

Dividing both equation by cosθ we get,

$3\frac{cosθ}{cosθ}−4\frac{sinθ}{cosθ}=2\frac{cosθ}{cosθ}+\frac{sinθ}{cosθ}$

3 – 4tanθ = 2 + tanθ

3 – 2 = 4tanθ + tanθ

tanθ = 1/5

问题 36. 如果∠A和∠B是锐角，使得tan A = tan B，则证明∠A = ∠B。

解决方案：

Let us consider a △ABC  

From the figure,

Given:

tan A = tan B

BC/AC = AC/BC

AC² = BC²

AC = BC

In △ABC, AC = BC So we can say that the triangle is an isosceles triangle, √3

and in an isosceles triangle we know that if two sides of a triangle are equal,  

then the angle opposite to the sides are equal.

Therefore ∠A =∠B