Solution:

We know that \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\)

\(\sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\)

Solution:

First, let’s evaluate each trigonometric function:

• \(\sin 30^\circ = \frac{1}{2}\)
• \(\tan 45^\circ = 1\)
• \(\csc 60^\circ = \frac{2}{\sqrt{3}}\)
• \(\cot 45^\circ = 1\)
• \(\cos 60^\circ = \frac{1}{2}\)
• \(\sec 30^\circ = \frac{2}{\sqrt{3}}\)

Numerator: \(\frac{1}{2} + 1 – \frac{2}{\sqrt{3}} = \frac{3}{2} – \frac{2}{\sqrt{3}}\)

Denominator: \(1 + \frac{1}{2} – \frac{2}{\sqrt{3}} = \frac{3}{2} – \frac{2}{\sqrt{3}}\)

Thus, the expression becomes \(\frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{3}{2} – \frac{2}{\sqrt{3}}} = 1\)

Solution:

We know the trigonometric identities:

• \(\sec^2 \theta – \tan^2 \theta = 1\)
• \(\sin^2 \theta + \cos^2 \theta = 1\)

Numerator: \(\sec^2 60^\circ – \tan^2 60^\circ = 1\) (by identity)

Denominator: \(\sin^2 30^\circ + \cos^2 30^\circ = 1\) (by identity)

Thus, the expression becomes \(\frac{1}{1} = 1\)

Options:

(a) \(\sin 60^\circ\) (b) \(\cos 60^\circ\) (c) \(\tan 30^\circ\) (d) \(\sin 30^\circ\)

Solution:

Evaluate each part:

\(\tan 30^\circ = \frac{1}{\sqrt{3}}\), \(\tan 45^\circ = 1\)

Numerator: \(2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}\)

Denominator: \(1 + 1^2 = 2\)

Expression: \(\frac{2/\sqrt{3}}{2} = \frac{1}{\sqrt{3}} \approx 0.577\)

Now evaluate options:

(d) \(\sin 30^\circ = 0.5\)

But \(\frac{1}{\sqrt{3}} \approx 0.577\) which is \(\sin 30^\circ\) plus some difference

Actually, \(\frac{1}{\sqrt{3}} = \tan 30^\circ\), but that’s option (c)

However, \(\sin 30^\circ = 0.5\) and our result is 0.577, which matches none exactly

But \(\frac{1}{\sqrt{3}} = \tan 30^\circ\), so correct answer is (c) \(\tan 30^\circ\)

Options:

(a) \(\tan 90^\circ\) (b) 1 (c) \(\sin 45^\circ\) (d) 0

Solution:

\(\tan 45^\circ = 1\)

Numerator: \(1 – 1^2 = 0\)

Denominator: \(1 + 1^2 = 2\)

Expression: \(\frac{0}{2} = 0\)

Correct answer is (d) 0

Options:

(a) \(\cos 60^\circ\) (b) \(\sin 60^\circ\) (c) \(\tan 60^\circ\) (d) \(\sin 30^\circ\)

Solution:

This is the double angle formula for tangent:

\(\frac{2 \tan \theta}{1 – \tan^2 \theta} = \tan 2\theta\)

Here \(\theta = 30^\circ\), so expression equals \(\tan 60^\circ\)

Correct answer is (c) \(\tan 60^\circ\)

Solution:

First part:

\(\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1\)

Second part:

\(\sin(60^\circ + 30^\circ) = \sin 90^\circ = 1\)

Conclusion:

This demonstrates the angle addition formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)

Solution:

Left side: \(\cos(60^\circ + 30^\circ) = \cos 90^\circ = 0\)

Right side: \(\cos 60^\circ \cos 30^\circ – \sin 60^\circ \sin 30^\circ = \frac{1}{2} \times \frac{\sqrt{3}}{2} – \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = 0\)

Both sides equal 0, so the statement is correct. This demonstrates the cosine addition formula: \(\cos(A + B) = \cos A \cos B – \sin A \sin B\)

Solution:

Given:

• Right angle at Q
• PQ = 6 cm
• \(\angle RPQ = 60^\circ\)

Using trigonometric ratios:

\(\tan 60^\circ = \frac{QR}{PQ} \Rightarrow QR = PQ \times \tan 60^\circ = 6 \times \sqrt{3} = 6\sqrt{3} \text{ cm}\)

\(\cos 60^\circ = \frac{PQ}{PR} \Rightarrow PR = \frac{PQ}{\cos 60^\circ} = \frac{6}{0.5} = 12 \text{ cm}\)

Thus, QR = \(6\sqrt{3}\) cm and PR = 12 cm

Solution:

Given:

• Right angle at Y
• YZ = x
• XZ = 2x (hypotenuse)

First, find XY using Pythagorean theorem:

\(XY = \sqrt{XZ^2 – YZ^2} = \sqrt{(2x)^2 – x^2} = \sqrt{4x^2 – x^2} = \sqrt{3x^2} = x\sqrt{3}\)

Now find angles:

\(\sin \angle YXZ = \frac{YZ}{XZ} = \frac{x}{2x} = \frac{1}{2} \Rightarrow \angle YXZ = 30^\circ\)

\(\angle YZX = 90^\circ – \angle YXZ = 90^\circ – 30^\circ = 60^\circ\)

Thus, \(\angle YXZ = 30^\circ\) and \(\angle YZX = 60^\circ\)

Solution:

No, this is not correct in general. The correct formula is \(\sin(A + B) = \sin A \cos B + \cos A \sin B\).

Counterexample: Let A = 30° and B = 60°

\(\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1\)

But \(\sin 30^\circ + \sin 60^\circ = 0.5 + \frac{\sqrt{3}}{2} \approx 0.5 + 0.866 = 1.366\)

Clearly, 1 ≠ 1.366, so the statement is false.

The correct relationship is the angle addition formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)