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Trigonometry Questions and Answers

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91.

The value of \(cosec(75°+\theta) - sec(15°-\theta) - tan(55°+\theta) + cot(35°-\theta)\) is:

Answer: B

We use the complementary angle identities: \(\sec(90°-A) = \csc A\) and \(\cot(90°-A) = \tan A\).

\(\sec(15°-\theta) = \sec(90° - (75°+\theta)) = \csc(75°+\theta)\).

\(\cot(35°-\theta) = \cot(90° - (55°+\theta)) = \tan(55°+\theta)\).

The expression becomes: \(\csc(75°+\theta) - \csc(75°+\theta) - \tan(55°+\theta) + \tan(55°+\theta) = 0\).

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92.

What is the value of \(\sin(420°)\)?

Answer: C

Subtract a full turn to bring the angle into range.

\[ 420^{\circ} - 360^{\circ} = 60^{\circ} \]

\[ \sin 60^{\circ} = \frac{\sqrt{3}}{2} \]

Sine has period 360°, so the value repeats exactly.

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93.

If \(\tan \theta = \frac{12}{5}\), what is the value of \(1+\sin\theta\)?

Answer: D

If \(\tan \theta = 12/5\) (Opposite/Adjacent), we find the hypotenuse: \(Hyp = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13\).

\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13}\).

The expression is \(1 + \frac{12}{13} = \frac{13+12}{13} = \frac{25}{13}\).

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94.

The expression \(\frac{\sin A}{1-\cot A} + \frac{\cos A}{1-\tan A}\) can be simplified to:

Answer: C

Convert to sin and cos: \(\frac{\sin A}{1-\frac{\cos A}{\sin A}} + \frac{\cos A}{1-\frac{\sin A}{\cos A}} = \frac{\sin^2 A}{\sin A - \cos A} + \frac{\cos^2 A}{\cos A - \sin A}\)

= \(\frac{\sin^2 A}{\sin A - \cos A} - \frac{\cos^2 A}{\sin A - \cos A}\)

= \(\frac{\sin^2 A - \cos^2 A}{\sin A - \cos A}\)

= \(\frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A - \cos A} = \sin A + \cos A\).

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95.

A ramp for a wheelchair has an angle of elevation of 5°. If the ramp is 12 feet long, how high off the ground is the top of the ramp? (Given sin 5° ≈ 0.087)

Answer: B

Let h be the height and L be the length of the ramp.

\(\sin(\text{angle}) = \frac{\text{height}}{\text{length}}\)

\(\sin 5° = \frac{h}{12}\)

h = 12 × \(\sin 5°\) ≈ 12 × 0.087 = 1.044 feet.

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96.

If \(A+B=90°\), then \(\sqrt{\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}}\) is equal to:

Answer: A

Since A+B=90°, we have B=90-A. So, \(\tan B = \cot A\), \(\cot B = \tan A\), \(\sec B = \csc A\), \(\sin B = \cos A\).

Substitute these into the expression under the root:

Numerator: \(\tan A \cot A + \tan A \tan A = 1 + \tan^2 A = \sec^2 A\).

Denominator: \(\sin A \csc A = 1\).

Second term: \(\frac{\cos^2 A}{\cos^2 A} = 1\).

The expression becomes \(\sqrt{\frac{\sec^2 A}{1} - 1} = \sqrt{\sec^2 A - 1}\).

Using the identity \(\sec^2 A - 1 = \tan^2 A\), we get \(\sqrt{\tan^2 A} = \tan A\).

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97.

The value of \(\tan 315°\) is:

Answer: B

315° lies in the fourth quadrant, where the tangent is negative.

\[ \tan 315^{\circ} = -\tan(360^{\circ} - 315^{\circ}) = -\tan 45^{\circ} = -1 \]

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98.

What is the value of \(\sin(-60°)\)?

Answer: B

Sine is an odd function.

\[ \sin(-\theta) = -\sin\theta \]

\[ \sin(-60^{\circ}) = -\frac{\sqrt{3}}{2} \]

Cosine, by contrast, is even: \( \cos(-\theta) = \cos\theta \).

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99.

The minimum value of \(\sec^2\theta + \csc^2\theta\) is:

Answer: C

We rewrite the expression using identities:

\(\sec^2\theta + \csc^2\theta = (1+\tan^2\theta) + (1+\cot^2\theta) = 2 + \tan^2\theta + \cot^2\theta\).

Let \(y = \tan^2\theta\). The expression is \(2 + y + 1/y\). By AM-GM inequality, \(y+1/y \ge 2\sqrt{y \cdot 1/y} = 2\). The minimum value of \(y+1/y\) is 2.

Therefore, the minimum value of the entire expression is 2 + 2 = 4.

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100.

The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. The tangent of one of its acute angles is:

Answer: D

Let the two legs of the right triangle be a and b. Hypotenuse c = 25 cm.

Perimeter = a + b + c = 60. So, a + b + 25 = 60 => a + b = 35.

By Pythagorean theorem, a² + b² = c² = 25² = 625.

We know \((a+b)^2 = a^2+b^2+2ab\). So, \(35^2 = 625 + 2ab \Rightarrow 1225 = 625 + 2ab \Rightarrow 2ab = 600 \Rightarrow ab = 300\).

We need two numbers that add to 35 and multiply to 300. The numbers are 15 and 20.

The legs are 15 cm and 20 cm. Let one acute angle be A. The tangent values would be opposite/adjacent, which could be 15/20 = 3/4 or 20/15 = 4/3.

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Tags: Trigonometry Questions and Answers || Trigonometry MCQ Questions || Trigonometric Functions Quantitative Aptitude || Trigonometry GK Questions || Quantitative Aptitude Trigonometry