InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
The horizontal distance between the two towers is \(\frac{{80}}{{\sqrt 3 }}m\) . The angle of depression at the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 250 m, find the height of the first tower.1). 180 m2). 150 m3). 223 m4). 160 m |
| Answer» | |
| 52. |
What is the simplified value of \(\sqrt {\frac{{\sec A}}{{\sec A - 1}} + \frac{{\sec A}}{{\sec A + 1}}?} \)1). cosec A2). √2 cosec A3). 2 sec2A4). sec A |
|
Answer» We can WRITE sec A = 1/ COS A, Now it can be written as ⇒ √{1/(1 - cos A) + 1/(1 + cos A) } ⇒ √[{(1 - cos A) + (1 + cos A) }/{(1 - cos A) (1 + cos A) }] ⇒ √{2/(1 - cos2A) } since we know that 1 - cos2 A = sin2 A ⇒ √2/sin A = √2 COSEC A ∴ Its simplified VALUE is √2 cosec A |
|
| 53. |
1). 193.222). 144.043). 176.124). 161.05 |
| Answer» | |
| 54. |
What is the simplified value of \(\sqrt {\frac{{sec\;A - 1\;}}{{sec\;A + 1\;}}} ?\)1). cosec A – cot A2). sec A – tan A3). sec 2 A4). sec A cosec A |
|
Answer» GIVEN: √[(secA – 1)/(secA +1)] ⇒ √[{(1/COSA) – 1}/{(1/cosA) +1}] ⇒ √[(1 - cosA)/(1 + cosA)] Multiplying (1 - cosA) on both numerator and denominator ⇒ √[(1 – cosA)2/(1 – cos2A)] ⇒ √[(1 – cosA)2/sin2A] ⇒ (1 – cosA)/sinA ⇒ 1/sinA – cosA/sinA ⇒ cosecA – cotA |
|
| 55. |
If sinA/√ (1 – sin2A) = x, then the value of x is?1). tanA2). cotA3). cosecA4). cosA |
|
Answer» ? 1 – sin2A = cos2A ⇒ sinA/√cos2A = x ⇒ sinA/cosA = x ⇒ TANA = x |
|
| 56. |
1). 2sinθcosθ2). cos\(\theta\)3). \(\sqrt 2 \)sin4). sin |
| Answer» | |
| 57. |
Find the value of (p + q)2, if p + q Cot3 α = Cot α.Cosec α and q = p tan α.1). 1 + Cos α2). 1 + Sin 2α3). 1 – cos 2α4). 1 - Sin α |
|
Answer» p + q COT3 α = Cot α.Cosec α ⇒ p + q COS3 α/Sin3 α = Cos α/Sin2 α Putting q = p TAN α in the equation we get ⇒ p + p tan α.(Cos3 α/Sin3 α) = Cos α/Sin2 α Putting tan α = sin α /cos α and simplifying we get ⇒ p(1 + Cos2 α/Sin2 α ) = Cos α/Sin2 α ⇒ p = Cos α Putting that in q = p tan α we get q = Sin α Here (p + q)2 = (Sin α + Cos α)2 Using (a + b)2 = a2 + b2 + 2AB we get (p + q)2 = Sin2 α + Cos2 α + 2 Sin α.Cos α Using Sin2 α + Cos2 α = 1 and 2 Sin α.Cos α = Sin 2α We get (p + q)2 = 1 + Sin 2α |
|
| 58. |
Find the Minimum Value of \({\bf{sec}}{\;^2}{\bf{x}}\; + \;{\bf{cosec}}{\;^2}{\bf{x}}\)1). 22). 43). 64). 8 |
|
Answer» $({\bf{sec}}{^2}{\bf{x}} + {\bf{cosec}}{^2}{\bf{x}} = 1{\rm{}} + {\rm{}}{\tan ^2}{\rm{x}} + {\rm{}}1{\rm{}} + {\rm{}}{\COT ^2}{\rm{x}} = {\rm{}}2{\rm{}} + {\rm{}}{\tan ^2}{\rm{x}} + {\rm{}}{\cot ^2}{\rm{x}})$ Now we have to find the minimum value of $(\begin{array}{l} \RIGHTARROW {\bf{ta}}{{\bf{n}}^2}{\bf{x}} + {\cot ^2}{\bf{x}} = {\LEFT( {\tan {\bf{x}}} \RIGHT)^2} + {\left( {\cot {\bf{x}}} \right)^2} - 2 \times \tan {\bf{x}} \times \\cot {\bf{x}} + 2 \times \tan {\bf{x}} \times \cot {\bf{x}}\\ \Rightarrow {\left( {\tan {\bf{x}} - \cot {\bf{x}}} \right)^2} + 2 = {\left( {\tan {\bf{x}} - \cot {\bf{x}}} \right)^2} + 2\end{array})$ For the value to be minimum, $(\tan {\bf{x}} - \cot {\bf{x}}\; = \;0)$ ⇒ Minimum value = 2 + 2 = 4 |
|
| 59. |
\(\frac{{\sqrt {1{\rm{\;}} + {\rm{\;sin\theta }}} }}{{\sqrt {1{\rm{\;}} - {\rm{\;sin\theta }}} }}\) is equal to1). secθ - tanθ2). cosecθ - cotθ3). secθ - cotθ4). secθ + tanθ |
|
Answer» $(\FRAC{{\SQRT {1{\rm{\;}} + {\rm{\;sin\THETA }}} }}{{\sqrt {1{\rm{\;}} - {\rm{\;sin\theta }}} }} = \frac{{\sqrt {1{\rm{\;}} + {\rm{\;sin\theta }}} }}{{\sqrt {1{\rm{\;}} - {\rm{\;sin\theta }}} }} \TIMES \frac{{\sqrt {1{\rm{\;}} + {\rm{\;sin\theta }}} }}{{\sqrt {1{\rm{\;}} + {\rm{\;sin\theta }}} }} = \frac{{\sqrt {\LEFT( {1{\rm{\;}} + {\rm{\;sin\theta }}} \right)2} }}{{\sqrt {1{\rm{\;}} - {\rm{\;sin}}2{\rm{\theta }}} }} = \frac{{\sqrt {\left( {1{\rm{\;}} + {\rm{\;sin\theta }}} \right)2} }}{{\sqrt {{\rm{cos}}2{\rm{\theta }}} }} = \frac{{1{\rm{\;}} + {\rm{\;sin\theta }}}}{{{\rm{cos\theta }}}} = \frac{1}{{{\rm{cos\theta }}}} + \frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}})$ |
|
| 60. |
1). \(\frac{{{h^2}\; + \;{x^2}}}{{4h}}\;metres\)2). \(\frac{{{h^2} - {x^2}}}{{4h}}\;metres\)3). \(\frac{{{h^2}\; + \;{x^2}}}{{2h}}\;metres\)4). \(\frac{{{h^2} - {x^2}}}{{2h}}\;metres\) |
| Answer» | |
| 61. |
If cos x = sin y and cot (x – 40°) = tan (50° –y), then the value of x and y are1). x = 70°, y = 20°2). x = 75°, y = 15°3). x = 85°, y = 5°4). x = 80°, y = 10° |
|
Answer» cos X = sin y ⇒ cos x = cos(90° – y) ⇒ x = 90° – y ⇒ x + y = 90°….(1) COT (x – 40°) = tan (50° – y) ⇒ cot (x – 40°) = cot (90° – (50° – y)) ⇒ x – 40° = 40° + y ⇒ x – y = 80°….(2) SOLVING (1) and (2) We get, x = 85° and y = 5° |
|
| 62. |
1). 32°2). 76°3). 25°4). 78° |
|
Answer» The SUM of all the interior angles of a triangle is 180° ⇒ ∠A + ∠B + ∠C = 180° ⇒ x + 4x – 4 + y = 180 Also, ∠C - ∠B = 8° ⇒ y – (4x - 4) = 8 ⇒ - 4x + y = 4- - - (2) On SOLVING equations (1) and (2) simultaneously, we get, x = 20° and y = 84 ° ∴ ∠B = (4 × 20) – 4 = 76° |
|
| 63. |
What is the value of \(\frac{{\left( {cos10^\circ + sin20^\circ } \right)}}{{\left( {cos20^\circ - sin10^\circ } \right)}}?\)1). \(\frac{1}{{\sqrt 3 }}\)2). \(- \frac{1}{{\sqrt 3 }}\)3). \(\sqrt 3\)4). \(- \sqrt 3\) |
|
Answer» Concept: $(\begin{ARRAY}{L} \COS \left( {90 - \theta } \right) = \sin \theta \\ \sin \left( {90 - \theta } \right) = \cos \theta \END{array})$ $(\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right))$ $(\sin A - \sin B = 2\sin \left( {\frac{{A - B}}{2}} \right)\cos \left( {\frac{{A + B}}{2}} \right))$ Calculation: $(\frac{{\cos 10^\circ + \sin 20^\circ }}{{\cos 20^\circ - \sin 10^\circ }} = \frac{{\cos \left( {90^\circ - 80^\circ } \right) + \sin 20^\circ }}{{\cos \left( {90^\circ - 70^\circ } \right) - \sin 10^\circ }})$ $(= \frac{{\sin 80^\circ + \sin 20^\circ }}{{\sin 70^\circ - \sin 10^\circ }} = \frac{{2\sin 50^\circ \cos 30^\circ }}{{2\cos 40^\circ \sin 30^\circ }})$ $(= \frac{{\sin \left( {90^\circ - 40^\circ } \right)\cot 30^\circ }}{{\cos 40^\circ }})$ $(= \frac{{\cos 40^\circ \cot 30^\circ }}{{\cos 40^\circ }} = \cot 30^\circ = \sqrt 3 )$ |
|
| 64. |
1). 02). 13). cosec θ4). sec θ |
|
Answer» [(sec2θ + 1)√(SEC$2$θ – 1)] × 1/2 × (cotθ – tanθ)$ ⇒$ [(sec2θ + 1)√(TAN$2$θ)] × 1/2 × (cosθ/sinθ – sinθ/cosθ)$ ⇒$ [(sec2θ + 1).tanθ] × 1/2 × [(cos2θ – sin2θ) / (sinθ.cosθ)]$ ⇒$ 1/2 × [(sec2θ + 1).tanθ] × [cos2θ/(sinθ.cosθ)]$ ⇒$ 1/2 × [(1 + cos2θ).tanθ] / (sinθ.cosθ)$ ⇒$ 1/2 × [(1 + 2cos2θ – 1) × sinθ/cosθ] / (sinθ.conθ)$ ⇒$ 1/2 × 2cos2θ × sinθ/cosθ × 1/(sinθ.cosθ)$ ⇒$ 1$ |
|
| 65. |
If 10tanA.tanB = 9, what is the value of cos(A – B)/cos(A + B)?1). 12). 03). 194). 9/10 |
|
Answer» We know, COS(A – B) = cosAcosB + sinAsinB And, cos(A + B) = cosAcosB – sinAsinB Given, tanAtanB = 9/10 cos(A – B)/cos(A + B) ⇒ (cosAcosB + sinAsinB)/(cosAcosB – sinAsinB) ⇒ (1 + tanAtanB)/(1 – tanAtanB) ⇒ (1+ 9/10)/(1 – 9/10) ⇒ 19 ∴ cos (A – B)/cos (A + B) = 19 |
|
| 66. |
1). 29/212). 21/293). 21/204). 20/29 |
|
Answer» SINCE we know that COS θ = √(1 - sin2 θ) ⇒ Cos θ = √(1 - 202/292) ⇒ Cos θ = √{(841 - 400)/292} ⇒ Cos θ = √{(212)/292} = 21/29 ∴ Cos θ = 21/29 |
|
| 67. |
In a right angled triangle if sinB = (5/13) then find the value of tanB + cosecB.1). 151/652). 181/603). 191/754). None of these |
|
Answer» GIVEN sinB = (5/13) = Perpendicular/hypotenuse ⇒ We know that (Hypotenuse)2 = (Perpendicular)2 + (BASE)2 ⇒ (13)2 = (5)2 + (Base)2 ⇒ (Base)2 = (13)2 - (5)2 ⇒ (Base)2 = 169 - 25 ⇒ (Base)2 = 144 ⇒ (Base) = √144 ⇒ (Base) = 12 ⇒ cosB = Base/hypotenuse = 12/13 ⇒ tanB = SIN B/Cos B ⇒ tanB = (5/13)/(12/13) ⇒ tanB = 5/12 ⇒ cosecB = (1/Sin B) = (13/5) ⇒ tanB + cosecB = (5/12) + (13/5) ⇒ (25 + 156)/60 ⇒ 181/60 ∴ Required value is 181/60 |
|
| 68. |
Find the value of (tan9° – tan27° – tan63° + tan81°).1). 12). 03). 34). 4 |
| Answer» | |
| 69. |
is 45°. If the length of the shadow increases by (√3 – 1)x, then the angle of elevation of the sun should become.1). 30°2). 45°3). 60°4). 75° |
| Answer» | |
| 70. |
The value of 4cot2 45° - sec230° is1). \(\frac{{ - 1}}{4}\)2). \(\frac{1}{4}\)3). \(\frac{8}{3}\)4). \(\frac{{ - 1}}{2}\) |
|
Answer» We know, ⇒ Cot 45 = 1 and sec 30 = 2/ √3 ⇒ ? = 4 × 1 - (2/ √3)2 ⇒ ? = 4 - 4/3 ⇒ ? = 8/3 |
|
| 71. |
What is the simplified value of \({\left( {\frac{1}{{cosecA\; + \;CotA}}} \right)^2}\)?1). sec A + tan A2). (1 - cos A)/(1 + cos A)3). (1 - cosec A)/(1 + cosec A)4). sin A |
|
Answer» COSEC A + Cot A = (1 + Cos A) /SIN A ⇒ 1/( Cosec A + Cot A)2 = (Sin A)2 /(1 + Cos A)2----(1) ⇒ (Sin A)2 = 1 - (COSA)2 = (1 + Cos A)(1 - CosA) PUTTING in (1) we get, ⇒ (1 - cos A)/(1 + cos A) |
|
| 72. |
What is the minimum value of sin2 θ + cos2 θ + sec2 θ + cosec2 θ + tan2 θ + cot2 θ1). 12). 33). 54). 7 |
|
Answer» We know that sin2 θ + cos2 θ = 1 Therefore, (sin2 θ + cos2 θ) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ ⇒ 1 + sec2 θ + cosec2 θ + tan2 θ + cot2 θ Using A.M ≥ G.M logic for tan2 θ + cot2 θ we get, ⇒ 1 + 2 + $({\sec ^2}{\rm{\theta \;}} + {\rm{\;cose}}{{\rm{c}}^2}{\rm{\;\theta }})$ Changing into sin and cos values (Because we know maximum and minimum values of Sin θ, Cos θ and by using simple identities we can convert all trigonometric functions into equation with SINE and Cosine.) $(\Rightarrow {\rm{\;}}1 + {\rm{\;}}2 + \left( {\frac{1}{{{{\cos }^2}{\rm{\theta }}}}} \right) + \left( {\frac{1}{{{{\sin }^2}{\rm{\theta }}}}} \right))$ Solving taking L.C.M $( \Rightarrow 1 + 2 + \frac{{{{\sin }^2}{\rm{\theta \;}} + {\rm{\;}}{{\cos }^2}{\rm{\theta }}}}{{{{\sin }^2}{\rm{\theta \;}}.{{\cos }^2}{\rm{\theta }}}})$ ------ Equation (1) But we already know two things Min. value of (sin θ cos θ)n = (½)n Apply them into Equation (1), and we get ⇒ 1 + 2 + (sin2 θ + cos2 θ)/( sin2 θ . cos2 θ) = 1 + 2 + (1/1/4) = 1 + 2 + 4 = 7 |
|
| 73. |
What is the simplified value of \({\left[ {\frac{{co{s^2}\theta }}{{1\; + \;\sin \theta }} - \frac{{si{n^2}\theta }}{{1\; + \;\cos \theta }}} \right]^2}\)?1). sinθ2). 1 – sin2θ3). 1 + sin2θ4). 1 – sinθ |
|
Answer» Since all the options have sinθ in them, we can CONCLUDE by seeing that they all give DIFFERENT VALUES. Putting θ = 45° We get the values of options as : (1) 1/√2 (2) 0 (3) 2 (4) (√2 – 1)/√2 Value asked in question = [(1/2)/(1 + (1/√2)) – (1/2)/(1 + (1/√2))]2 = 0 ∴ the answer is (2) |
|
| 74. |
1). 12). 03). 34). 4 |
|
Answer» ⇒ 2ycos θ = x sin θ----(1) ⇒ Now SQUARING on both side ⇒ 4y2cos2 θ = x2 sin2 θ----(2) ⇒ 2xsec θ – y COSEC θ = 3 ⇒ 2x/cos θ – y/sin θ = 3 ⇒ 2x sin θ – y cos θ = 3sin θcos θ ⇒ from equation 1 put ycos θ = xsin θ/2 ⇒ 2x sin θ – x sin θ/2 = 3 sin θ cos θ ⇒ 3 sin θ = 6 sin θ cos θ ⇒ x = 2cos θ----(3) ⇒ Now, x2 + 4y2 ⇒ (2cos θ)2 + x2 sin2 θ/cos² θ ⇒ 4cos2 θ + 4 cos2 θ sin2 θ/cos2 θ ∴ 4(cos2 θ + sin2 θ) = 4 |
|
| 75. |
A pole broken by the storm of wind and its top struck the ground at an angle of 30° and at a distance of 20 m from the foot of the pole. The height of the pole before it was broken was1). 20 √3 m2). \(\frac{{40\sqrt 3 }}{3}m\)3). 60 √3 m4). \(\frac{{100\sqrt 3 }}{3}m\) |
| Answer» | |
| 76. |
1). 45°2). 60°3). 15°4). 30° |
|
Answer» ⇒ tan (A + B) = √3 ⇒ tan (A + B) = tan 60° ⇒ A + B = 60°------ 1 ⇒ tan (A - B) = 1/√3 ⇒ tan (A - B) = tan 30° ⇒ A – B = 30°------ 2 ⇒ Adding equation 1 and 2 we get ⇒ 2A = 90° ⇒ A = 45° ⇒ PUTTING the VALUE of A in equation 1 we get ⇒ A + B = 60° ⇒ 45° + B = 60° ∴ ∠B = 15° |
|
| 77. |
If sec θ + tan θ = 2, then sec θ is equal to1). \(\frac{7}{4}\)2). \(\frac{7}{2}\)3). \(\frac{5}{2}\)4). \(\frac{5}{4}\) |
|
Answer» We KNOW that, Sec2 θ – tan2 θ = 1 Using (a2 – B2) = (a – b)(a +b), we GET (Sec θ + tan θ)(Sec θ - tan θ) = 1 Given, sec θ + tan θ = 2 . . . . . . . eq (i) ⇒ (Sec θ – tan θ) = 1/2 ....... eq (ii) Now, adding eq (i) and (ii), we get ⇒ 2 Sec θ = 2 + ½ ⇒ 2Sec θ = 5/2 ⇒ secθ = 5/4 |
|
| 78. |
If \(tan\theta = \frac{p}{q}\), then the value of \(\frac{{{\rm{psec\theta }} - {\rm{qcosec\theta }}}}{{{\rm{psec\theta }} + {\rm{qcosec\theta }}}}\) -1). \(\frac{{{\rm{p}} - {\rm{q}}}}{{{\rm{p}} + {\rm{q}}}}\)2). \(\frac{{{{\rm{q}}^2} - {{\rm{p}}^2}}}{{ - {{\rm{q}}^2} + {{\rm{p}}^2}}}\)3). \(\frac{{{{\rm{p}}^2} - {{\rm{q}}^2}}}{{{{\rm{q}}^2} + {{\rm{p}}^2}}}\)4). 1 |
|
Answer» <P>$(\FRAC{{{\rm{psec\theta }} - {\rm{qcosec\theta }}}}{{{\rm{psec\theta }} + {\rm{cosec\theta }}}} = \frac{{\frac{p}{{cos\theta }} - \frac{Q}{{sin\theta }}}}{{\frac{p}{{cos\theta }} + \frac{q}{{sin\theta }}}})$ $(= \frac{{ptan\theta- q}}{{ptan\theta + q}})$(by MULTIPLYING sinθ) $(= \frac{{p \times \frac{p}{q} - q}}{{p \times \frac{p}{q} + q}} = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}})$ |
|
| 79. |
Maximum value of (6sinθ + 8cosθ) is –1). 142). 13). 104). 28 |
|
Answer» We know that, Maximum value of,(a sinθ + b cosθ) is √(a2 + b2) ∴ maximum value of (6sinθ + 8cosθ) is = √(62 + 82) = √(36 + 64) = √(100) = 10 |
|
| 80. |
What is the value of (2sec 45o + tan 30o)?1). (2√6 + 1)/√32). √33). (2√2 + 3)/√64). (9 + 2√3)/9 |
|
Answer» Since, we know that SEC 45 = √2 and tan 30 = 1/√3, PUT this in the REQUIRED equation ⇒ 2 sec 45o + tan 30o = 2 × √2 + 1/√3 = (2√6 + 1)/√3 ∴ The VALUE of 2 sec 45o + tan 30o is (2√6 + 1)/√3 |
|
| 81. |
What is the simplified value of (cosA + sinA)(cotA + tanA)?1). secA + cosecA2). sinA + cosA3). tanA + cotA4). secA – cosecA |
|
Answer» GIVEN: (cosA + SINA)(cotA + tanA) ⇒ (cosA + sinA)[(cosA) / (sinA) + (sinA) / (cosA)] ⇒ (cosA + sinA)[(cos2A) + (sin2A) / (sinA)(cosA)] ⇒ (cosA + sinA)[1/ (sinA)(cosA)] ⇒ 1/sinA + 1/cosA ⇒ cosecA + SECA |
|
| 82. |
In a triangle ABC, ∠A = y, ∠B = x and ∠C = y + 20. If 2x – y = 20, then the triangles is:1). Right-angled2). Obtuse-angled3). Equilateral4). Acute angled |
|
Answer» SUM of angles = 180° y + x + y + 20 = 180 x + 2y = 160----(1) 2x – y = 20----(2) From (1) and (2) x = 40° and y = 60° The angles are 40°, 60° and 80° The triangle is an ACUTE ANGLED triangle |
|
| 83. |
Find the Minimum value of 8 tan 2x + 7 cot 2x1). 3√562). 2√563). √564). 8√28 |
|
Answer» We have, $(8\;{\bf{ta}}{{\bf{N}}^2}{\bf{x}}\; + \;7{\cot ^2}{\bf{x}}\; = \;{\left( {\sqrt 8 \tan {\bf{x}}} \right)^2}\; + \;{\left( {\sqrt 7 \cot {\bf{x}}} \right)^2} - 2\; \times \;\sqrt 8 \tan {\bf{x}}\; \times \;\sqrt 7 \cot {\bf{x}}\; + \;2\; \times \;\sqrt 8 \tan {\bf{x}}\; \times \;\sqrt 7 \cot {\bf{x}})$ $(\Rightarrow \;{\left( {\sqrt 8 \tan {\bf{x}} - \sqrt 7 \cot {\bf{x}}} \right)^2}\; + \;2\sqrt {8\; \times \;7} \; = \;{\left( {\sqrt 8 \tan {\bf{x}} - \sqrt 7 \cot {\bf{x}}} \right)^2}\; + \;2\sqrt {56} )$ For the value to be minimum, $(\sqrt 8 \tan {\bf{x}} - \sqrt 7 \cot {\bf{x}}\; = \;0)$ ⇒ Minimum value = 2√56 |
|
| 84. |
If cos x = sin y and cot (x - 35°) = tan (35° - y), then the value of x and y are1). x = 90°, y = 0°2). x = 75°, y = 15°3). x = 85°, y = 5°4). x = 80°, y = 10° |
|
Answer» cos x = sin y ⇒ cos x = cos(90° – y) ⇒ x = 90° – y ⇒ x + y = 90°….(1) cot (x - 35°) = TAN (35° - y) ⇒ cot (x – 35°) = cot (90° – (35° - y)) ⇒ x – 35° = 55° + y ⇒ x – y = 90°….(2) Solving (1) and (2) We GET, x = 90° and y = 0° |
|
| 85. |
If sinθ + cosθ = √3 cos (90 - θ), then what is the value of tanθ?1). √3 - 12). √3 + 13). (√3 + 1)/24). (√3 - 1)/2 |
|
Answer» sinθ + cosθ = √3 sinθ cosθ = √3 sinθ - sinθ cosθ = sinθ (√3 - 1) ∴ tanθ = 1/(√3 - 1) = (√3 + 1)/2 |
|
| 86. |
Value of the expression \(\frac{{3{\rm{sin}}72^\circ }}{{{\rm{cos}}18^\circ }} - \frac{{{\rm{sec}}32^\circ }}{{{\rm{cosec}}58}}\) is1). -22). 13). 24). 3 |
|
Answer» $(\FRAC{{3{\rm{sin}}72^\circ }}{{{\rm{COS}}18^\circ }} - \frac{{{\rm{sec}}32^\circ }}{{{\rm{cosec}}58}})$ $( \Rightarrow \frac{{3{\rm{sin}}\left( {90^\circ {\rm{\;}} - {\rm{\;}}18^\circ } \right)}}{{{\rm{cos}}18^\circ }} - \frac{{{\rm{sec}}\left( {90^\circ {\rm{\;}} - {\rm{\;}}58^\circ } \right)}}{{{\rm{cosec}}58^\circ }} = \frac{{3\cos 18^\circ }}{{{\rm{cos}}18^\circ }} - \frac{{{\rm{cosec}}58^\circ }}{{{\rm{cosec}}58^\circ }})$ ⇒ 3 × 1 - 1 = 2 |
|
| 87. |
If cot2A = x, then x is1). cosec2A - 12). (cosec2A + 1)/2cotA3). (cot2A + 1)/2cotA4). 2cotA/(1 + cot2A) |
|
Answer» X = cot2A since cot2A = cos2A/sin2A = (1 – sin2A)/sin2A = cosec2A – 1 |
|
| 89. |
What is the value of cot 11π/6?1). -√32). -13). √34). -1/2 |
|
Answer» ⇒ COT(11π/6) ⇒ cot(2π – π/6) ⇒ -cot(π/6)? cot is NEGATIVE in FOURTH quadrant. ∴ - √3 |
|
| 90. |
If A = 30°, B = 60° and C = 135°, then what is the value of sin3A + cos3B + tan3C – 3sin A cos B tan C?1). 02). 13). 84). 9 |
|
Answer» To find : sin3(30°) + cos3(60°) + tan3(135°) – 3sin (30°) cos (60°) tan (135°) ⇒ (1/2) 3 + (1/2) 3 + (-1) 3 – 3 X (1/2) x (1/2) x (-1) ⇒ 1/8 + 1/8 -1 + 3/4 ⇒ 1/4 – 1 + 3/4 ⇒ 0 |
|
| 91. |
What is the simplified value of \(\sqrt {\frac{{(cosec\;A)}}{{(cosec\;A) - 1}} + \frac{{(cosec\;A)}}{{(cosec\;A) + 1}}} \)1). √(2) secA2). √(2cosecA)3). sec 2A4). 2secA |
|
Answer» Given: $(\sqrt {\FRAC{{cosec\;A}}{{cosec\;A - 1}} + \frac{{cosec\;A}}{{cosec\;A + 1}}?} )$ Substituting cosecA = 1/sinA ⇒ √{[(1/sinA) / (1/sinA – 1)] + [1/sinA/ (1/sinA + 1)]} ⇒ √{[1/(1 – sinA)] + [1/(1 + sinA)]} ⇒ √{[1 + sinA + 1 – sinA]/[1 – sin2A]} ⇒ √{[2/[1 – sin2A]} ⇒ √{[2/cos2A} ⇒ √{[2sec2A} ⇒ √(2) secA ∴ the CORRECT option is 1) |
|
| 92. |
What is the value of (tan30o + √3/2)?1). (4 + √3)/22). 5/2√3 3). (4 + √3)/2√34). (√3 + 1)/√3 |
| Answer» | |
| 93. |
What is the value of tan 6° tan 36° tan 84° tan 54° tan 45°?1). 1/22). 1/√23). 14). 1/3 |
|
Answer» TAN 6° tan 36° tan 84° tan 54° tan 45° ⇒ tan 6° tan 36° tan (90° - 6°) tan (90° - 36°) tan 45° ⇒ tan 6° tan 36° COT 6° cot 36° tan 45°[? tan (90 - θ) = cot θ] ⇒ tan 6° tan 36° × (1/tan 6°) × (1/tan 36°) tan 45° ⇒ tan 45° ⇒ 1 |
|
| 94. |
If tanθ.cos60° = √3/2, then the value of sin (θ – 15°) is1). √3/22). 1/23). 14). 1/√2 |
| Answer» | |
| 95. |
Find the minimum value of (asec2 θ + bcosec2 θ).1). (a + b)2). (a + b) + 2√ab3). 2√ab4). 1 |
|
Answer» The given FUNCTION can be written as? a(1 + tan2 θ) + b(1 + cot2 θ) = a + b + atan2 θ + bcot2 θ = (a + b) + (atan2 θ + bcot2 θ) Now, formula for FINDING min value of (atan2 θ + bcot2 θ) = 2√ab ∴ Minimum value of the given expression = (a + b) + 2√ab |
|
| 96. |
If 7 sin2θ + 3 cos2θ = 4, and θ is a positive acute angle, then tan θ is equal to 1). 1/32). 1/73). \(\frac{1}{{\sqrt 3 }}\)4). \(\sqrt 3\) |
|
Answer» We know, SIN2θ + cos2θ = 1 Given, 7 sin2θ + 3 cos2θ = 4 ⇒ 4sin2θ + 3(sin2θ + cos2θ) = 4 ⇒ 4sin2θ + 3 = 4 ⇒ sin2θ = ¼ ⇒ sinθ = ½ or -1/2 Given, θ is a POSITIVE ACUTE angle ∴ sinθ = 1/2 As, sin 30° = ½ ∴ sinθ = sin 30° ⇒ θ = 30° We have to find the value of tanθ = tan30° = 1/√3 |
|
| 97. |
The angles of elevation of the top of a tower 72 metre high from the top and bottom of a building are 30° and 60° respectively. What is the height (in metres) of building?1). 422). 20√33). 24√34). 48 |
| Answer» | |
| 98. |
1). 12). 7/153). 3/54). 0 |
|
Answer» ⇒ Let x = $(\frac{{\SQRT {1 - sin{\RM{\THETA }}} }}{{\sqrt {1 + sin{\rm{\theta }}} }})$ ⇒ Multiplying NUMERATORAND denominator by √(1 - sinθ) we get ⇒ x = (1 - sinθ)/√(1 - sin2 θ) ⇒ x = (1 - sinθ)/√(cos2 θ) ⇒ x = SEC θ - tan θ whentan θ = (8/15) according to Pythagorean theorem: Sec θ = 17/15 so, ⇒ x = (17/15) - (8/15) ∴ x = 9/15 = 3/5 |
|
| 99. |
If Sin θ = 12/13, then what is the value of Cot θ?1). 13/122). 5/133). 5/124). 13/5 |
|
Answer» GIVEN SIN θ = 12/13, so Cosec θ = 13/12 COT θ can be given as ⇒ Cot θ = √(cosec2 θ - 1) = √(132/122 - 1) = √{(169 - 144) /122} = √(52/122) = 5/12 ∴ Cot θ is 5/12 |
|
| 100. |
(Assume that ψ and θ are both positive acute angles with ψ < 75° and θ < 60°)1). 12). √33). 1/√34). √3/2 |
|
Answer» ⇒ 75° - ψ = 60° - θ = 45° ⇒ ψ = 30°; θ = 15° ∴ COT(θ + ψ) = cot 45° = 1 |
|