10类RD Sharma解决方案–第5章三角比–练习5.1 |套装2

问题7.如果cotθ= 7/8，请评估：

(一世) $\frac{(1+sinθ)(1-sinθ)}{(1+cosθ)(1-cosθ)}$

(ii)婴儿床^2θ

解决方案：

 cotθ = 7/8 = Base/Perpendicular

In right-angled ΔPQR,

∠Q = 90°, PQ = 8, RQ = 7

Using Pythagoras Theorem

PR² = PQ² + QR²

PR² = 8² + 7² = 64 + 49

PR² = 113

PR = √113
Now

sinθ = Perpendicular/Hypotenuse = PQ/PR = 8/√113     

cosθ = Base/Hypotenuse = QR/PR = 7/√113            

(i) $\frac{(1+sinθ)(1-sinθ)}{(1+cosθ)(1-cosθ)}$

Putting the values of sinθ and cosθ in the equation

$=\frac{(1+\frac{8}{\sqrt113})(1-\frac{8}{\sqrt113})}{(1+\frac{7}{\sqrt113} )(1-\frac{7}{\sqrt113} )}$

=$\frac{(1)^2-(\frac{8}{\sqrt113})^2}{(1)^2-(\frac{7}{\sqrt113})^2}$

=$\frac{1-\frac{64}{113}}{1-\frac{49}{113}}$

$=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}$

$=\frac{\frac{49}{113}}{\frac{64}{113}}$

= 49/64

(ii) cot²θ  

= (cosθ/sinθ)² 

Putting the values of sinθ and cosθ in the following equation

=$[\frac{(\frac{7}{\sqrt113})}{(\frac{8}{\sqrt113})}]^2$

=$\frac{(\frac{49}{113})}{(\frac{64}{113})}$

= 49/64

or

cot²θ = (cotθ)² = (7/8)² = 49/64

问题8。如果3 cot A = 4，请检查是否$\frac{1−tan^2A}{1+tan^2A} = cos^2 A - sin^2 A$或不。

解决方案：

Given, 3cot A = 4 or cot A = 4/3

Draw a △ ABC where ∠B = 90°, AB = 4, BC = 3

Using Pythagoras Theorem

AC² = AB² + BC²

AC² = 4² + 3² = 16 + 9

AC² = 25

AC = 5

Now, 

Taking LHS

=$\frac{1−tan^2A}{1+tan^2A}$

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

=$\frac{1−(\frac{9}{16})}{1+(\frac{9}{16})}$

=$\frac{(\frac{16-9}{16})}{(\frac{16+9}{16})}$ 

= 7/25

Taking RHS

= cos²A – sin²A

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

=($\frac{16}{25})-(\frac{9}{25})$

=($\frac{16-9}{25})$

= 7/25

RHS = LHS (Hence Proved)

问题9。如果tanθ= a / b，则求出值$\frac{(cosθ+sinθ)}{(cosθ-sinθ)}$

解决方案：

Given, tanθ = a/b

Draw a △ABC where ∠B = 90°, AB = b, BC = a

Using Pythagoras Theorem

AC² = BC² + AB²

AC² = a² + b²

AC² = $\sqrt{(a^2+b^2)}$

Now, 

$\frac{(cosθ+sinθ)}{(cosθ-sinθ)}$

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

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

= (b + a)/(b – a)

问题10。如果3tanθ= 4，则求出值$\frac{(4cosθ-sinθ)}{(2cosθ+sinθ)}$

解决方案：

Given tanθ = 4/3 

Now, Dividing the numerator and denominator by cosθ 

=$\frac{(4-tanθ)}{(2+tanθ)}$

Putting the values of tanθ in the above equation

=$\frac{(4-\frac{4}{3})}{(2+\frac{4}{3})}$

=$\frac{(\frac{16-3}{4})}{(\frac{8+3}{4})}$

= 8/10

= 4/5

问题11。如果3cotθ= 2，则求出值$\frac{(4sinθ-3cosθ)}{(2sinθ+6cosθ)}$

解决方案：

Given: 3cotθ = 2

  Using Pythagoras Theorem

AC² = BC² + AB²

AC² = 3² + 4²

AC² = 9 + 16 = 25

AC = 5

Now, 

=$\frac{(4sinθ-3cosθ)}{(2sinθ+6cosθ)}$ 

$=\frac{(4\frac{3}{\sqrt13}-3\frac{2}{\sqrt13})}{(2\frac{3}{\sqrt13}+6\frac{2}{\sqrt13})}$

$=\frac{(\frac{12}{\sqrt13}-\frac{6}{\sqrt13})}{(\frac{6}{\sqrt13}+\frac{12}{\sqrt13})}$ 

$=\frac{\frac{12-6}{\sqrt13}}{\frac{6+12}{\sqrt13}}$ 

$=\frac{\frac{6}{\sqrt13}}{\frac{18}{\sqrt13}}$

= 6/18 = 1/3 

问题12。如果tanθ= a / b，则证明$\frac{(asinθ-bcosθ)}{(asinθ+bcosθ)}=\frac{a^2-b^2}{a^2+b^2}$

解决方案：

Given, tanθ = a/b 

  

Using Pythagoras Theorem

AC² = BC² + AB²

AC² = a² + b²

AC² = $\sqrt{(a^2+b^2)}$

Now,   

=$\frac{(asinθ-bcosθ)}{(asinθ+bcosθ)}$

Putting the values of sinθ and cosθ in the above equation

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

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

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

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

Hence Proved

问题13。如果secθ= 13/5，则证明$\frac{(2sinθ-3cosθ)}{(4sinθ-9cosθ)}$ = 3

解决方案：

Given, secθ = 13/5

Using Pythagoras Theorem

AC² = BC² + AB²

13² = BC² + 5²

BC² = 169 – 25 = 144

BC = 12

Now,    

Taking LHS 

=$\frac{(2sinθ-3cosθ)}{(4sinθ-9cosθ)}$

Putting the values of sinθ and cosθ in the above equation

$=\frac{(2\frac{12}{13}-3\frac{5}{13})}{(4\frac{12}{13}-9\frac{5}{13})}$

$=\frac{(\frac{24}{13}-\frac{15}{13})}{(\frac{48}{13}-\frac{45}{13})}$

$=\frac{(\frac{24-15}{13})}{(\frac{48-45}{13})}$

$=\frac{(\frac{9}{13})}{(\frac{3}{13})}$ 

= 3 = RHS

Hence Proved

问题14.如果cosθ= 12/13，则表明sinθ(1 –tanθ)= 35/156

解决方案：

We have cosθ = 12/13

Using Pythagoras Theorem

AC² = BC² + AB²

13² = BC² + 12²

BC² = 169 – 144 = 25

BC = 5

Now, 

Taking LHS

= sinθ(1 – tanθ)

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

=$\frac{5}{13}.\frac{7}{12}$

= 35/156

= RHS

Hence Proved

问题15.如果cotθ= 1 /√3，则表明$\frac{1−cos^2θ}{ 2−sin^2θ}=\frac{3}{5}$

解决方案：

Given, cotθ = 1/√3

tanθ = 1/cotθ =√3 

From Pythagoras theorem,

AC² = AB² + BC²

AC² = 1² + (√3)²

AC² = 3 + 1 = 4

AC = 2

Now, 

Taking LHS

=$\frac{1−cos^2θ}{ 2−sin^2θ}$

=$\frac{1-(\frac{1}{2})^2}{2-(\frac{\sqrt3}{2})^2}$

$=\frac{(4−1)}{(8−3)}$

= 3/5         

Hence Proved

问题16。如果tanθ= 1 /√7，则$\frac{(cosec^2θ-sec^2θ)}{(cosec^2θ+sec^2θ)}=\frac{3}{4}$

解决方案：

We have

tanθ = 1/√7

cotθ = √7

We know sec²θ = (1 + tan²θ) = 1 + 1/7 = 8/7

and cosec²θ = (1 + cot²θ) = 1 + 7 = 8

Now, 

$\frac{(cosec^2θ-sec^2θ)}{(cosec^2θ+sec^2θ)}$

=$\frac{(8-\frac{8}{7})}{(8+\frac{8}{7})}$

= 48/64 = 3/4

问题17。如果secθ= 5/4，则求出值$\frac{sinθ-2cosθ}{tanθ-cotθ}$

解决方案：

Given:

secθ = 5/4

cosθ = 1/secθ = 4/5

From Pythagoras theorem,

AC² = BC² + AB²

5² = BC² + 4²

BC² = 25 − 16 = 9

BC = 3

Now,

=$\frac{sinθ−2cosθ}{tanθ−cotθ}=\frac{\frac{3}{5}-2(\frac{4}{5})}{\frac{3}{4}-\frac{4}{3}}$

=$\frac{\frac{-5}{5}}{\frac{-7}{12}}$

= 12/7   

问题18。如果tanθ= 12/13，则求出值$\frac{2sinθcosθ}{cos^2θ−sin^2θ}$

解决方案：

Given: tanθ = 12/13

From Pythagoras theorem,

AC² = BC² + AB²

AC² = (13)² + (12)²

AC² = 313

AC = √313

sinθ = 12/√313

cosθ = 13/√313

We have

$\frac{2sinθcosθ}{cos^2θ−sin^2θ} = \frac{2\times\frac{12}{\sqrt{133}} \times \frac{13}{\sqrt{133}}}{(\frac{12}{\sqrt{133}})^2 - (\frac{13}{\sqrt{133}})^2 }$

= \$\frac{\frac{312}{313}}{\frac{169}{313}-\frac{144}{313}}$

= $\frac{\frac{312}{313}}{\frac{25}{313}}$

= 312/25

问题19.如果cosθ= 3/5，则求值$\frac{sinθ−\frac{1}{tanθ}}{2tanθ}$

解决方案：

Given:

cosθ = 3/5

From Pythagoras theorem,

AC² = BC² + AB²

5² = 3² + AB²

AB² = 25 − 9 = 16

AB = 4

Now

$\frac{sinθ−\frac{1}{tanθ}}{2tanθ}$

=$\frac{\frac{4}{5}-\frac{1}{\frac{4}{3}}}{2\frac{4}{3}}$

=$\frac{\frac{4}{5}-\frac{3}{4}}{\frac{8}{3}}$

=$\frac{\frac{16-15}{20}}{\frac{8}{3}}$

= (1/20) × (3/8)

= 3/160

问题20 。如果sinθ= 3/5，则求值$\frac{cosθ−\frac{1}{tanθ}}{2cotθ}$

解决方案：

Given,

sinθ = 3/5

Now

$\frac{sinθ−\frac{1}{tanθ}}{2cotθ}$

=$\frac{sinθ−{cotθ}}{2cotθ}$

=$\frac{cosθ−\frac{cosθ}{sinθ}}{2\frac{cosθ}{sinθ}}$

=$\frac{cosθ−\frac{cosθ}{sinθ}}{2\frac{cosθ}{sinθ}}$

=$\frac{cosθ}{cosθ}.\frac{1−\frac{1}{sinθ}}{\frac{2}{sinθ}}$

=$\frac{\frac{sinθ-1}{sinθ}}{\frac{2}{sinθ}}$

= (sinθ – 1)/(2)

Putting the value of sinθ, we get

=$\frac{\frac{3}{5}-1}{2}$ 

=$\frac{3-5}{10}$

= -1/5  

问题21.如果tanθ= 24/7，则求出sinθ+cosθ。

解决方案：

Given:   

tanθ = 24/7 

From Pythagoras theorem,

AC² = BC² + AB²

AC² = 24² + 7²

AC² = 576 + 49 = 625

AC = 25

Now, 

= sinθ + cosθ

= 24/25 + 7/25

= (24 + 7)/25

= 31/25