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Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.2 | Set 2
  • Last Updated : 21 Feb, 2021

Evaluate each of the following(14-19)

Question 14. \frac{sin30°-sin90°+2cos0°}{tan30°tan60°}     

Solution: 

Given:\frac{sin30°-sin90°+2cos0°}{tan30°tan60°}           -(1)

Putting the values of sin 30° = 1/2, tan 30° = 1/√3, tan 60° = √3, sin 90° = cos 0° = 1 in eq(1) 

\frac{(\frac{1}{2})-(1)+2(1)}{(\sqrt3)(\frac{1}{\sqrt3})}

\frac{(\frac{1}{2})+1}{1}



= 3/2 

Question 15. \frac{1}{cot^230°}+\frac{1}{sin^260°}−cos^245°

Solution:  

Given: \frac{4}{cot^230°}+\frac{1}{sin^260°}−cos^245°           -(1)

= tan230° + cosec260° − cos245°

Putting the values of cosec60° = 2/√3, cos45° = 1/√2, tan30° = 1/√3 in eq(1)  

(\frac{4}{\sqrt3})^2+(\frac{2}{\sqrt3})^2+(\frac{1}{\sqrt2})^2

= 4/3 + 4/3 – 1/2

= (4 + 4)/3 – 1/2



= 8/3 – 1/2

= 13/6

Question 16. 4(sin430° + cos260°) – 3(cos245° – sin290°) – sin260°

Solution:

Given: 4(sin430° + cos260°) – 3(cos245° – sin290°) – sin260°       -(1)

Putting the values of cos45° = 1/√2, sin30° = cos30° = 1/2, sin60° = √3/2, sin90° = 1 in eq(1)

= 4((1/2)4 + (1/2)2) – 3((1/√2)2 – (1)2) – (√3/2)2 

= 4((1/16) + (1/4)) – 3((1/2) – 1) – (3/4)

= 4(5/16) – 3(-1/2) – 3/4

= 5/4 + 3/2 – 3/4

= (5 + 6 – 3)/4

= 8/4

= 2

Question 17. \frac{tan^260°+4cos^245°+3sec^230°+5cos^290°}{cosec30°+sec60°−cot^230°}

Solution:

Given: \frac{tan^260°+4cos^245°+3sec^230°+5cos^290°}{cosec30°+sec60°−cot^230°}

=\frac{(\sqrt3)^2+4(\frac{1}{\sqrt2})^2+3(\frac{2}{\sqrt3})^2+5(0)}{2+2−(\sqrt3)^2}

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

= 3 + 4(1/2) + 3(4/3)

= 3 + 2 + 4

= 9

Question 18.  Evaluate \frac{sin30°}{sin45°}+\frac{tan45°}{sec60°}-\frac{sin60°}{cot45°}-\frac{cos30°}{sin90°}

Solution: 

Given: \frac{sin30°}{sin45°}+\frac{tan45°}{sec60°}-\frac{sin60°}{cot45°}-\frac{cos30°}{sin90°}

\frac{(\frac{1}{2})}{(\frac{1}{\sqrt2})}+\frac{1}{2}-\frac{\frac{\sqrt3}{2}}{1}-\frac{\frac{\sqrt3}{2}}{1}

\frac{\sqrt2}{2}+\frac{1}{2}-\frac{\sqrt3}{2}-\frac{\sqrt3}{2}

\frac{\sqrt2+1-2\sqrt3}{2}         

Question 19. \frac{tan45°}{cosec30°}+\frac{sec60°}{cot45°}-\frac{5sin90°}{2cos0°}

Solution:

Given: \frac{tan45°}{cosec30°}+\frac{sec60°}{cot45°}-\frac{5sin90°}{2cos0°}

As we know that 

cosecθ = 1/sinθ, cotθ = 1/tanθ

Now,

= tan45°sin30° + sec60°tan45° – \frac{5sin90°}{2cos0°}

= 1 – 1/2 + 2 – 5/2

= (1 – 5)/2 + 2

= -2 + 2

= 0 

Find the value of X in each of the following 

Question 20. 2sin3x = √3

Solution:

Given,

2sin3x = √3

sin3x = √3/2

As we know that,

sin60° = √3/2

So, sin3x = sin60°

Now, 

3x = 60°

x = 60°/3

x = 20°

Question 21. 2sin x/2

Solution:

Given

2sinx/2 = 1 or sin x/2 = 1/2

We know sin30° = 1/2

So, 

sinx/2 = sin30°

x/2 = 30°

x = 60°

Question 22. √3sinx = cosx

Solution:

Given,

√3sinx = cosx or sinx/cosx = 1/√3

tan x = 1/√3

We know tan30° = 1/√3

So,

tan x = tan30°

x = 30°

Question 23. tan x = sin45°cos45° + sin30°

Solution:

Given: tan x = sin45°cos45° + sin30°

Putting the values of sin 45° = cos 45° = 1/√2, sin 30° = 1/2

tan x = ((1/√2 × 1/√2) + 1/2) 

tan x = (1/√2)2 + 1/2

tan x = 1/2 + 1/2

tan x = 1

tan x = tan45°

So, x = 45°

Question 24. √3tan 2x = cos60° + sin45°cos45°

Solution:

Putting the values of sin 45° = cos 45° = 1/√2, cos 60° = 1/2 in the given equation  

√3tan 2x = 1/2 + 1/√2 × 1/√2

√3tan 2x = 1/2 + (1/√2)2

√3tan 2x = 1/2 + 1/2

tan 2x = 1/√3

tan 2x = tan30°

So, x = 15°

Question 25. cos 2x = cos60°cos30° + sin60°sin30°

Solution:

Given: cos 2x = cos60°cos30° + sin60°sin30°

Putting the values of sin 30° = cos 60° = 1/2, cos 30° = sin 60° = √3/2

cos 2x = 1/2 × √3/2 + √3/2 × 1/2

cos 2x = √3/4 + √3/4

cos 2x = √3/2

cos 2x = cos30°

So, x = 15°

Question 26. If θ = 30°, verify that:

(i) tan2θ = \frac{2tanθ}{1−tan^2θ}

Solution:

Putting the value of given θ in the above equation

We get

tan2(30°)=\frac{2tan30°}{1−tan^230°}

tan 60°=\frac{2(\frac{1}{\sqrt3})}{1−(\frac{1}{\sqrt3})^2}

√3 = \frac{\frac{2}{\sqrt3}}{1−\frac{1}{3}}

√3 = \frac{\frac{2}{\sqrt3}}{\frac{2}{3}}

√3 = √3

Hence Proved

(ii) tan2θ = \frac{4tanθ}{1+tan^2θ}

Solution:

Putting the value of given θ in the above equation

We get

tan2(30°)=\frac{4tan30°}{1+tan^230°}

tan 60°=\frac{4(\frac{1}{\sqrt3})}{1+(\frac{1}{\sqrt3})^2}

√3 = \frac{\frac{4}{\sqrt3}}{1+\frac{1}{3}}

√3 = \frac{\frac{4}{\sqrt3}}{\frac{4}{3}}

√3 = √3 

Hence Proved

(iii) cos2θ = \frac{1−tan^2θ}{1+tan^2θ}

Solution:

Putting the value of given θ in the above equation

We get

cos2(30°)=\frac{1-tan^230°}{1+tan^230°}

cos60°=\frac{1-(\frac{1}{\sqrt3})^2}{1+(\frac{1}{\sqrt3})^2}

1/2=(3-1)(3+1)

1/2 = 2/4 or 1/2

Hence Proved

(iv) cos3θ = 4cos3θ − 3cosθ

Solution:

Putting the value of given θ in the above equation

We get

cos3(30°) = 4cos3 30° − 3cos 30°

cos 90° = 4cos330° − 3cos 30°

0=4(\frac{\sqrt3}{2})^3-3(\frac{\sqrt3}{2})

0 = 4(3√3/8) – 4(3√3/8)

0 = 3(√3/2) – 3(√3/2)

0 = 0

Hence Proved

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