Class 10 NCERT解决方案-第8章三角学入门–练习8.4

问题1.用cot A表示三角比sin A，sec A和tan A

解决方案：

(i) sin A

We know that

cosec²A = 1 + cot²A

1/sin²A = 1 + cot²A

sin²A = 1/(1 + cot²A)

sin A = 1/(1+cot²A)^1/2

(ii) sec A 

sec²A = 1 + tan²A

Sec²A = 1 + 1/cot²A

sec²A = (cot²A + 1) / cot²A

sec A = (cot²A + 1)^1/2 / cot A

(iii) tan A

tan A = 1 / cot A

tan A = cot ^-1 A

问题2.用sec A来写∠A的所有其他三角比。

解决方案：

(i) cos A

cos A = 1/sec A

(ii) sin A

We know that

sin²A = 1 – cos²A

Also , cos²A = 1 / sec²A

sin²A = 1 – 1 / sec²A

sin²A = (sec²A – 1) / sec²A

sin A = (sec²A – 1)^1/2 / sec A

(iii) tan A

We know that

tan²A + 1 = sec²A

tan A = (sec²A – 1)½

(iv) cosec A

We know

cosec A = 1/ sinA

cosec A = sec A / (sec²A – 1)½

(v) cot A

We know

cot A = cos A / sin A

cot A = (1/sec A) / ((sec²A – 1)^1/2 / sec A)

cot A = 1 / (sec²A – 1)^1/2

问题3.评估：

(i)(正弦² 63°+正弦² 27°)/(余弦² 17°+余弦² 73°)

(ii)正弦25°cos 65°+余弦25°sin 65°

(i) ([sin(90-27)]² + sin² 27) / ([cos(90-73)]² + cos² 73)

We know that 

sin(90-x) = cos x
cos(90-x) = sin x

(cos²(27) + sin² 27) / (sin²(73) + cos² 73)

Using 

sin²A + cos²A = 1

1/1 = 1

(ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]

Using

sin(90-x) = cos x
cos(90-x) = sin x

= [sin 25 * sin 25] + [cos 25 * cos 25]

= sin² 25 + cos² 25

= 1

问题4.选择正确的选项。证明您的选择。

解决方案：

(i) 9 sec² A – 9 tan² A  

(A) 1 (B) 9 (C) 8 (D) 0 

Using sec²A – tan²A = 1 

9 (sec²A – tan²A ) = 9(1) 

Ans (B) 

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

(A) 0 (B) 1 (C) 2 (D) –1 

Simplifying all ratios

= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)

= ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ)

= ((cosθ + sinθ)² – 1) / (sinθ cosθ)

= (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ)

= 2

Ans (C)

(iii) (sec A + tan A) * (1 – sin A) 

(A) sec A (B) sin A (C) cosec A (D) cos A 

Simplifying sec A and tan A

= (1/cos A + sin A/cos A)*(1 – sin A)

= ((1 + sin A)/cos A)*(1 – sin A)

= (1 – sin²A)/cos A

= cos²A / cos A

= cos A

Ans (D)

(iv) (1 + tan²A) / (1 + cot²A)

(A) sec²A (B) –1 (C) cot²A (D) tan²A

Simplifying tan A and cot A

= (1 + (sin²A / cos²A)) / (1 + (cos²A / sin²A))

= ((cos²A + sin²A) / cos²A) / ((cos²A + sin²A) / sin²A)

= sin²A / cos²A

= tan²A

Ans (D)

问题5.证明以下身份，其中涉及的角度是定义了表达式的锐角。

解决方案：

(i) (cosec θ – cot θ)² = (1 – cosθ) / (1 + cosθ)

Solving LHS

Simplifying cosec θ and cot θ

= (1-cos θ)² / sin²θ

= (1-cos θ)² / (1-cos²θ)

Using a² – b² = (a+b)*(a-b)

= (1-cos θ)² / [(1-cos θ)*(1+cos θ)]

= (1-cos θ) / (1+cos θ) = RHS

Hence Proved

(ii) (cos A / (1+sin A) + ((1+sin A) / cos A) = 2 sec A

Solving LHS

Taking LCM 

= (cos²A + (1+sin A)²) / ((1+sin A) cos A)

= (cos²A + 1 + sin²A + 2 sin A ) / ((1 + sin A)*cos A)

Using sin²A + cos²A = 1

= (2 + 2*sin A) / ((1+sin A)*cos A)

= (2*(1 + sin A)) / ((1 + sin A)*cos A)

= 2 / cos A

= 2 sec A = RHS

Hence Proved

(iii) (tan θ / (1 – cot θ)) + (cot θ / (1 – tan θ)) = 1 + sec θ*cosec θ

Solving LHS

Changing tan θ and cot θ in terms of sin θ and cos θ and simplifying

= ((sin²θ) / (cos θ *(sin θ-cos θ))) + ((cos²θ ) / (sin θ *(sin θ-cos θ)))

= (1 / (sin θ-cos θ)) * [(sin³θ – cos³θ) / (sin θ * cos θ)]

= (1 / (sin θ – cos θ)) * [ ((sin θ – cos θ) * ( sin²θ + cos²θ + sin θ * cos θ ))/(sin θ *cos θ)]

= (1+sin θ*cos θ) / (sin θ*cos θ)

= sec θ*cosec θ + 1 = RHS

Hence Proved

(iv) (1 + sec A) / sec A = sin²A / (1 – cos A)

Solving LHS

= cos A + 1

Solving RHS

= (1 – cos²A) / (1 – cos A)

= (1 – cos A) * (1 + cos A) / (1 – cos A)

= 1 + cos A = RHS

Hence Proved

(v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec²A = 1 + cot²A

Solving LHS

Multiplying numerator and denominator by (cot A – 1 + cosec A)

= (cot²A + 1 + cosec²A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot²A – (1 + cosec²A – 2*cosec A))

= (2*cosec²A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot²A – 1 – cosec²A + 2*cosec A) 

= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot²A – 1 – cosec²A + 2*cosec A)

= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2) 

= cosec A + cot A = RHS

Hence Proved

(vi) [(1 + sin A) / (1 – sin A)]^½ = sec A + tan A

Solving LHS

Multiplying numerator and denominator by (1+sinA)

= [((1 + sin A)*(1 + sin A)) / ((1 – sin A)*(1 + sin A))]^½

= (1 + sin A) / (1 – sin²A)^½

= (1 + sin A) / (cos²A)^1/2

= (1 + sin A) / (cos A)

= sec A + tan A = RHS

Hence Proved

(vii) (sin θ – 2 sin³θ) / (2 cos³θ – cos θ) = tan θ

Solving LHS

= (sin θ * (1 – 2*sin²θ)) / (cos θ * (2*cos²θ – 1))

= (sin θ * (1 – 2*sin²θ )) / (cos θ * (2*(1 – sin²θ) – 1))

= (sin θ *(1 – 2*sin²θ)) / (cos θ * (1 – 2*sin²θ))

= tan θ = RHS

Hence Proved 

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A

Solving LHS

= sin²A + cosec²A + 2*sin A *cosec A + cos²A + sec²A + 2*cos A *sec A

We know that cosec A = 1 / sin A

= 1 + 1 + cot²A + 1 + tan²A + 2 + 2

= 7 + tan²A + cot²A = RHS

Hence Proved

(ix) (cosec A – sin A)*(sec A – cos A) = 1 / (tan A + cot A)

Solving LHS

= ((1/sin A) – sin A) * ((1/cos A) – cos A)

= ((1 – sin²A) / sin A) * ((1 – cos²A) / cos A)

= (cos²A * sin²A) / (sin A * cos A)

= sin A * cos A

Solving RHS

Simplifying tan A and cot A

= (sin A * cos A) / ( sin²A + cos²A)

= sin A * cos A = RHS

Hence Proved

(x) (1 + tan²A) / (1 + cot²A ) = [(1 – tan A) / (1 – cot A)]² = tan²A

Solving LHS

Changing cot A = 1 / tan A

= (tan²A * (1 + tan²A)) / (1 + tan²A) = tan²A = RHS

= [(1 – tan A) / (1 – cot A)]² = (1 + tan²A – 2*tan A) / (1 + cot²A – 2*cot A)

= (sec²A – 2*tan A) / (cosec²A – 2*cot A)

Solving this we get 

= tan²A

Hence Proved