Practice MCQ Questions and Answer on Trigonometry

211.

The expression (tanθ + cotθ)(secθ + tanθ)(1 - sinθ), 0° θ 90° is equal to:

• (A) sinθ
• (B) secθ
• (C) cosecθ
• (D) cotθ

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 $$\eqalign{ & \left( {\tan \theta + \cot \theta } \right)\left( {\sec \theta + \tan \theta } \right)(1 - \sin \theta ) \cr & = \left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)\left( {1 - \sin \theta } \right) \cr & = \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)\left( {1 - \sin \theta } \right) \cr & = \left( {\frac{1}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{1 - {{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {\frac{1}{{\cos \theta \sin \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta }}{{\cos \theta }}} \right) \cr & = \frac{1}{{\sin \theta }} \cr & = {\text{cosec}}\,\theta \cr} $$

212.

What is the value of $$\frac{{\left[1-\tan \left({90}^{\circ }-\theta \right)\right]}^2}{\left[{\cos }^2\left({90}^{\circ }-\theta \right)\right]}-1=?$$

• (A) -sin2θ
• (B) -cos2θ
• (C) cos2θ
• (D) sin2θ

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 $$\eqalign{ & \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - \theta } \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - \theta } \right)} \right]}} - 1 \cr & {\text{By putting }}\theta = {45^ \circ } \cr & \Rightarrow \frac{{{{\left[ {1 - \tan \left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}^2}}}{{\left[ {{{\cos }^2}\left( {{{90}^ \circ } - {{45}^ \circ }} \right)} \right]}} - 1 \cr & \Rightarrow \frac{{{{\left[ {1 - \tan {{45}^ \circ }} \right]}^2}}}{{{{\cos }^2}{{45}^ \circ }}} - 1 \cr & \Rightarrow 0 - 1 \cr & \Rightarrow - 1 \cr & {\text{By satisfying in option A}} \cr & \Rightarrow - \sin 2\theta \cr & \Rightarrow - \sin {90^ \circ } \cr & \Rightarrow - 1 \cr} $$

213.

$${\left(\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3-\cos \theta }\right)}^2+1,$$     θ â‰Â  45° is equal to:

• (A) cosec2θ
• (B) sec2θ
• (C) cot2θ
• (D) 2tan2θ

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 $$\eqalign{ & {\left( {\frac{{\sin \theta - 2{{\sin }^3}\theta }}{{2{{\cos }^3} - \cos \theta }}} \right)^2} + 1 \cr & = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{\left( {\frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}} \right)^2} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

214.

Find $$\cos \left(-\frac{7\pi }{2}\right)=?$$

• (A) $$\frac{1}{2}$$
• (B) 1
• (C) -1
• (D) 0

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 $$\eqalign{ & \cos \left( { - \frac{{7\pi }}{2}} \right) \cr & = \cos 7 \times \frac{{180}}{2} \cr & = \cos 630 \cr & = \cos \left( {2 \times 360 - 90} \right) \cr & = \cos 90 \cr & = 0 \cr} $$

215.

$${\left(\frac{1-\tan \theta }{1-\cot \theta }\right)}^2+1=?$$

• (A) cosec2θ
• (B) sec2θ
• (C) cos2θ
• (D) sin2θ

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 $$\eqalign{ & {\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{1 - \tan \theta }}{{\frac{{\tan \theta - 1}}{{\tan \theta }}}}} \right)^2} + 1 \cr & = \frac{{\left( {1 - \tan \theta } \right){{\tan }^2}\theta }}{{\left( {1 - \tan \theta } \right)}} + 1 \cr & = {\tan ^2}\theta + 1 \cr & = {\sec ^2}\theta \cr} $$

216.

The value of $$\frac{1}{\sqrt{2}}\text{sin}\frac{\pi }{6}$$ . $$\text{cos}\frac{\pi }{4}$$ - $$\cot \frac{\pi }{3}$$ . $$\text{sec}\frac{\pi }{6}$$ + $$\frac{5\tan \frac{\pi }{4}}{12\sin \frac{\pi }{2}}$$   is equal to?

• (A) 0
• (B) 1
• (C) 2
• (D) $$\frac{3}{2}$$

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 $$\eqalign{ & \frac{1}{{\sqrt 2 }}{\text{sin}}\frac{\pi }{6}{\text{.cos}}\frac{\pi }{4} - \cot \frac{\pi }{3}{\text{.sec}}\frac{\pi }{6}{\text{ + }}\frac{{5\tan \frac{\pi }{4}}}{{12\sin \frac{\pi }{2}}} \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} \times \frac{2}{{\sqrt 3 }} + \frac{{5 \times 1}}{{12 \times 1}} \cr & \Rightarrow \frac{1}{4} - \frac{2}{3} + \frac{5}{{12}} \cr & \Rightarrow \frac{{3 - 8 + 5}}{{12}} \cr & \Rightarrow 0 \cr & {\text{ }} \cr} $$

217.

If sec²ÃƒÂƒÃ‚ŽÃ‚¸ + tan²ÃƒÂƒÃ‚ŽÃ‚¸ = $$3\frac{1}{2},$$  0° θ 90°, then (cosθ + sinθ) is equal to

• (A) $$\frac{1+\sqrt{5}}{3}$$
• (B) $$\frac{2+\sqrt{5}}{3}$$
• (C) $$\frac{1+\sqrt{5}}{6}$$
• (D) $$\frac{9+2\sqrt{5}}{6}$$

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 $$\eqalign{ & {\sec ^2}\theta + {\tan ^2}\theta = 3\frac{1}{2} \cr & \Rightarrow 1 + {\tan ^2}\theta + {\tan ^2}\theta = \frac{7}{2} \cr & \Rightarrow 2{\tan ^2}\theta = \frac{7}{2} - 1 \cr & \Rightarrow 2{\tan ^2}\theta = \frac{5}{2} \cr & \Rightarrow 3{\tan ^2}\theta = \frac{5}{4} \cr & \Rightarrow \tan \theta = \frac{{\sqrt 5 \to P}}{{2 \to B}} \cr & H = \sqrt {5 + 4} = 3 \cr & \therefore \,\cos \theta + \sin \theta \cr & = \frac{2}{3} + \frac{{\sqrt 5 }}{3} \cr & = \frac{{2 + \sqrt 5 }}{3} \cr} $$

218.

The value of sin²65° + sin²25° + cos²35° + cos²55° is?

• (A) 0
• (B) 1
• (C) 2
• (D) $$\frac{1}{2}$$

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 $$\eqalign{ & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}{25^ \circ } + {\cos ^2}{35^ \circ } + {\cos ^2}{55^ \circ } \cr & \Rightarrow {\sin ^2}{65^ \circ } + {\sin ^2}\left( {{{90}^ \circ } - {{65}^ \circ }} \right) + \left[ {{{\cos }^2}{{35}^ \circ } + {{\cos }^2}\left( {{{90}^ \circ } - {{35}^ \circ }} \right)} \right] \cr & \Rightarrow \left( {{{\sin }^2}{{65}^ \circ } + {{\cos }^2}{{65}^ \circ }} \right) + \left( {{{\cos }^2}{{35}^ \circ } + si{n^2}{{35}^ \circ }} \right) \cr & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr} $$

219.

If rsinθ = 1, rcosθ = $$\sqrt{3}\text{,}$$  then the value of r²tanθ is?

• (A) 4
• (B) $$\frac{1}{\sqrt{3}}$$
• (C) $$\frac{4}{\sqrt{3}}$$
• (D) $$\text{4}\sqrt{3}$$

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 $$\eqalign{ & r\sin \theta = 1,{\text{ }}r\cos \theta = \sqrt 3 \cr & {\text{Put }}\theta = {30^ \circ } \cr & r = 2 \cr & {\text{So}},{\text{ }}{r^2}\tan \theta \cr & = {\left( 2 \right)^2} \times {\text{tan}}{30^ \circ } \cr & = 4 \times \frac{1}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

220.

In ΔABC, âˆÂ B = 90° and AB : BC = 2 : 1, then value of (sinA + cotC) = ?

• (A) $$3+\sqrt{5}$$
• (B) $$\frac{2+\sqrt{5}}{2\sqrt{5}}$$
• (C) $$2+\sqrt{5}$$
• (D) $$3\sqrt{5}$$

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 $$\eqalign{ & {\text{AC}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & {\text{sin A}} + \operatorname{cotC} \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} + \frac{{{\text{BC}}}}{{{\text{AB}}}} \cr & \frac{1}{{\sqrt 5 }} + \frac{1}{2} \cr & \Rightarrow \frac{{2 + \sqrt 5 }}{{2\sqrt 5 }} \cr} $$