Practice MCQ Questions and Answer on Trigonometry

31.

Maximum value of (2sinθ + 3cosθ) is?

• (A) 2
• (B) $$\sqrt{13}$$
• (C) $$\sqrt{15}$$
• (D) 1

Show Answer

 $$\eqalign{ & \left( {2\sin \theta + 3\cos \theta } \right) \cr & {\text{Maximum value of}} \cr & a\sin \theta + b\cos \theta \cr & = \sqrt {{a^2} + {b^2}} \cr & = \sqrt {{2^2} + {3^2}} \cr & = \sqrt {4 + 9} \cr & = \sqrt {13} \cr} $$

32.

If tan40° = α, then find $$\frac{\tan {320}^{\circ }-\tan {310}^{\circ }}{1+\tan {320}^{\circ }\cdot \tan {310}^{\circ }}$$

• (A) $$\frac{1-{\alpha }^2}{\alpha }$$
• (B) $$\frac{1+{\alpha }^2}{2\alpha }$$
• (C) $$\frac{1-{\alpha }^2}{2\alpha }$$
• (D) $$\frac{1+{\alpha }^2}{\alpha }$$

Show Answer

 $$\eqalign{ & \frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ }\tan {{310}^ \circ }}} \cr & = \frac{{\tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right) - \tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}}{{1 + \tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right)\tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}} \cr & = \frac{{ - \tan {{40}^ \circ } + \cot {{40}^ \circ }}}{{1 + \tan {{40}^ \circ } \times \cot {{40}^ \circ }}} \cr & = \frac{{\cot {{40}^ \circ } - \tan {{40}^ \circ }}}{{1 + 1}} \cr & = \frac{{\frac{1}{\alpha } - \alpha }}{2} \cr & = \frac{{1 - {\alpha ^2}}}{{2\alpha }} \cr} $$

33.

If sinθ × cosθ = $$\frac{1}{2}\text{,}$$ Then the value of sinθ - cosθ is where 0° θ 90°.

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

Show Answer

 $$\eqalign{ & \sin \theta \times \cos \theta = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & 2\sin \theta \times \cos \theta = 1 \cr & \sin 2\theta = 1 \cr & \sin 2\theta = \sin {90^ \circ } \cr & 2\theta = {90^ \circ } \cr & \theta = {45^ \circ } \cr & {\text{So,}}\sin \theta - {\text{cos}}\theta \cr & = \sin {45^ \circ } - \cos {45^ \circ } \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0{\text{ }} \cr} $$

34.

The value of cos²20° + cos²70° is?

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

Show Answer

 $$\eqalign{ & {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\sin ^2}{70^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = 1 \cr} $$

35.

If $$\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }=\frac{\text{k}+1}{\text{k}-1},$$     then k = ?

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

Show Answer

 $$\eqalign{ & \frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{{\text{k}} + 1}}{{{\text{k}} - 1}} \cr & \frac{{\tan \theta }}{{\sin \theta }} = {\text{k}} \cr & {\text{k}} = \sec \theta \cr} $$

36.

If 7sin²ÃƒÂƒÃ‚ŽÃ‚¸ + 3cos²ÃƒÂƒÃ‚ŽÃ‚¸ = 4, then the value of secθ + cosecθ is?

• (A) $$\frac{2}{\sqrt{3}}-2$$
• (B) $$\frac{2}{\sqrt{3}}+2$$
• (C) $$\frac{2}{\sqrt{3}}$$
• (D) None of these

Show Answer

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & 4{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{4} \cr & {\text{sin }}\theta = \frac{1}{2} \cr & \sin \theta = \sin {30^ \circ } \cr & \theta = {30^ \circ } \cr & \sec \theta + \operatorname{cosec} \theta \cr & = \sec {30^ \circ } + \operatorname{cosec} {30^ \circ } \cr & = \frac{2}{{\sqrt 3 }} + 2 \cr} $$

37.

If $$\pi \sin \theta =1,$$   $$\pi \cos \theta =1\text{,}$$   then the value of $$\left\{\sqrt{3}\tan \left(\frac{2}{3}\theta \right)+1\right\}$$     is?

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

Show Answer

 $$\eqalign{ & \pi \sin \theta = 1\,.....(i) \cr & \pi \cos \theta = 1\,.....(ii) \cr & {\text{Divide eq}}{\text{. (i) from (ii)}} \cr & \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr & \tan \theta = 1 \cr & \tan \theta = \tan {45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr & = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr & = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr & = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr & = 2 \cr} $$

38.

If $$\text{tan }\alpha =2,$$   then the value of $$\frac{\text{cose}{\text{c}}^2\alpha -\text{se}{\text{c}}^2\alpha }{\text{cose}{\text{c}}^2\alpha +se{c}^2\alpha }$$   is?

• (A) $$-\frac{5}{9}$$
• (B) $$\frac{3}{5}$$
• (C) $$-\frac{3}{5}$$
• (D) $$\frac{17}{5}$$

Show Answer

 $$\eqalign{ & {\text{tan}}\alpha = 2\left( {{\text{given}}} \right) \cr & \therefore \frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }} \cr} $$ (Divide by coses2α both in N and D) $$\eqalign{ & = \frac{{1 - {\text{ta}}{{\text{n}}^2}\alpha }}{{1 + {\text{ta}}{{\text{n}}^2}\alpha }} \cr & = \frac{{1 - {{\left( 2 \right)}^2}}}{{1 + {{\left( 2 \right)}^2}}} \cr & = - \frac{3}{5} \cr} $$

39.

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

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

Show Answer

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

40.

If secθ + tanθ = p, (p > 1) then $$\frac{\text{cosec}\theta +1}{\text{cosec}\theta -1}=?$$

• (A) 2p2
• (B) $$\frac{p+1}{p-1}$$
• (C) p2
• (D) $$\frac{p-1}{p+1}$$

Show Answer

 $$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$