# Trigonometry Questions and Answers for Competitive Exams Like CGL, CHSL, MTS at Smartkeeda

Directions : Read the following questions carefully and choose the right answer.
Important for :
1
Find the value of sin2 10 + sin2 20 + sin2 30 + ....... + sin2 80.
» Explain it
D
We can rewrite above equation as

sin2 10 + sin2 80 + sin2 20 + sin2 70 + sin2 30 + sin2 60 + sin2 40 + sin2 50   ….. equation (A)

We know that sin2 x + sin2 (90 – x) = 1

Therefore equation A becomes

1 + 1 + 1 + 1 = 4

Hence, option D is correct.

2
Find the value of 16/ √3 (Cos50°Cos10°Cos110°Cos60°)
» Explain it
C
We have cos cos (60 – ) cos (60 + )

= cos (cos cos 60 + Sin x Sin 60°) (cos cos 60° – Sin x Sin 60°)

= cos (cos2cos260° – Sin2Sin260°)

= cos x { (1 / 4) cos2 – (3 / 4) sin2 x }

= 1/4 {cos3 x – 3 cos x (1 – cos2 x)}

= 1/4 (4 cos3 x – 3 cos x)

= 1/4 cos 3 x

Thus,

cos x º cos (60 – x)º cos (60 + x) = 1/4 cos (3x)

Therefore,

cos 50º cos 10º cos 110º = (1/4) cos 150º

= 1/4 {–(√3)/2} = – (√3)/8 ...eq A

Also cos 60º = 1/2 ...eq B

Put values of eq.A and Eq.B in

16/(√3) (cos 50º cos 10 cos 110º cos 60º), we get

= 16/(√3) × {–(√3) / 8} × 1/2

= –1

Option C is hence the correct answer.

3
Find the value of

cos2 Θ {√(1 + sin Θ / 1 – sin Θ) + √(1 – sing Θ / 1 + sin Θ)}
» Explain it
C
cos2 θ [√{(1 + sin θ) (1 + sin θ) / (1 – sin θ) (1 + sin θ)} + √{(1 – sin θ) (1 – sin θ) / (1 + sin θ) (1 – sin θ)}]

⇒ cos2 θ [√{(1 + sin θ)2 / (1–sin2 θ)} + √{(1 – sin θ)2 / (1 – sin2 θ)}]

⇒ cos2 θ [√{(1 + sin θ)2 / cos2 θ} + √{(1 – sin θ)2 / cos2 θ}]

⇒ cos2 θ [{(1 + sin θ) / cos θ} + {(1 – sin θ) / cos θ}]

⇒ cos2 θ {(1 + sin θ + 1 – sin θ) / cos θ}

⇒ 2 cos2 θ / cos θ = 2 cos θ

Hence, option C is correct.

4
If sin 21° = x/y, then sec 21° – sin 69° is equal to
» Explain it
A
Sin 21º = x / y

Cos 21°= √(1 – (sin 21°)2 )

⇒ √{1 – (x2 / y2)} = {√(y2 –x2) / y}

⇒ sec 21º = { y / √(y2 – x2)}

According to the question,

⇒ sec 21°– sin⁡ 69°

⇒ sec⁡ 21°– sin⁡ (90– 21°)

⇒ sec⁡ 21° – cos⁡21°

⇒ { y / √(y2 –x2) – √(y2 –x2) / y }

⇒ { x2 / y √(y2 – x2)

Hence, option A is correct.

5
Find the value of :

(sin 35º / cos 55º)2 + (cos 55º / sin 35º)2 – 2 cos 30º
» Explain it
C
(sin 35º / cos 55º)2 + (cos 55º / sin 35º)2 – 2 cos 30º

⇒ {sin (90–35º) / cos 55º}2 + { cos (90 – 55º) / sin 35º}2 – 2 cos 30º

⇒ (cos 55º / cos 55º)2 + (sin 35º / sin 35º)2 – 2 cos 30º

⇒ 1 + 1 – 2 × {(√3) / 2}

⇒ 2 – √3

Hence, option C is correct.

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### Trigonometry Questions for Competitive Exams with PDF like SSC CGL Tier 1, CHSL, MTS, CGL Tier 2

Is it possible to find the height of a tower by merely observing the length of its shadow and the position of the sun? Is it within our powers to deduce the dimensions of a cylindrical container by measuring the height and circumference of the heap of grain that has been poured out of it? One doesn’t need to be a Sherlock Holmes to crack these mysteries. Anyone with an elementary knowledge of geometry will be able to answer these questions. Geometry plays a major role in SSC Exams 2 to 3 questions generally being asked in SSC CGL Tier 1 and SSC CHSL and questions number increased as per the total number of question in particular exams like in SSC CGL Tier 2, that value increased by 5 to 10 questions.
With geometry, one enters, quite literally, into the real world – the world of dimensions. We are not dealing with abstract equations or mere numbers but grappling with concepts like size, shape, location, direction and orientation of objects that figure prominently in real life.

Coordinate geometry and trigonometry find wide applications in varied fields like aviation, defense, navigation and prospecting. Many abstract designs that modern artists and architects create, and we so admire, are nothing more than creatively put together collections of geometric shapes. The concepts of area and perimeter are vital not just to people dealing in real estate but to anyone planning to make optimum use of any available space. As you can see, there is hardly any aspect of the real world that is not influenced by geometry.

At Smartkeeda you familiarized yourself with the key concepts and improved your problems-solving abilities. Practice and tests are important to optimize your preparation. Take the tests on Testzone to improve your problem-solving skills. Questions that have appeared in the previous SSC CGL and SSC CHSL exams are also in the given quizzes. Through them you can understand where the focus lies in an examination environment. The detailed solutions and short tricks of the questions may provide some alternate strategies that can help you improve your speed accuracy. You can practice individual exercise practice and you can analyze yourself topic wise as Trigonometry, Height and Distance, Quadrilateral and Polygon, Triangle, Circle, and Lines and Angles. Here are some Important Geometry Chapters Links:

Trigonometry Questions For SSC CGL Click Here...

Triangle Questions for SSC CGL Click Here...

Lines and Angles Questions for SSC CGL Click Here...

Height and Distance Questions for SSC CGL Click here...

I hope this above article and exercise will help you to crack your SSC CGL 2019 and SSC CHSL 2019 as well.
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