Q: What is the value of $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$?

We know that $\tan 45^\circ = 1$. \nSubstituting this value into the given expression: $$\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$$ \nHence, the correct option is (D).

Q: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$. Show all steps.

Given: $\sin A = \frac{3}{4}$ \nWe know that in a right-angled triangle, $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$. \nLet the perpendicular be $3k$ and the hypotenuse be $4k$, where $k$ is a positive real number.\nUsing Pythagoras theo

Q: Prove the trigonometric identity: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$. Show all steps clearly.

To prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ \nTake the Left Hand Side (LHS): $\text{LHS} = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$ \nFactor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator: $\te

Q: Prove the identity: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$. Show all steps clearly.

To prove: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$ \nTake the Left Hand Side (LHS): $\text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$ \nRationalize the denominator inside the square root by multiplying both numerator and denominator by $(1 + \sin A)$: $\text{LHS} = \sqrt{\frac{(1 +

Question 1

What is the value of $\frac{1 - 	an^2 45^\circ}{1 + 	an^2 45^\circ}$?

Accepted Answer

We know that $	an 45^\circ = 1$. 
Substituting this value into the given expression: $$\frac{1 - 	an^2 45^\circ}{1 + 	an^2 45^\circ} = \frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$$ 
Hence, the correct option is (D).

Question 2

If $\sin A = \frac{3}{4}$, then what is the value of $\cos A$?

Accepted Answer

Using the trigonometric identity $\sin^2 A + \cos^2 A = 1$: $$\cos^2 A = 1 - \sin^2 A$$ $$\cos^2 A = 1 - \left(\frac{3}{4}ight)^2 = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16}$$ 
Taking the square root on both sides: $$\cos A = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$ 
Hence, the corre

Question 3

If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $	an A$. Show all steps.

Accepted Answer

Given: $\sin A = \frac{3}{4}$ 
We know that in a right-angled triangle, $\sin A = \frac{	ext{Opposite side}}{	ext{Hypotenuse}} = \frac{	ext{Perpendicular}}{	ext{Hypotenuse}}$. 
Let the perpendicular be $3k$ and the hypotenuse be $4k$, where $k$ is a positive real number.
Using Pythagoras theo

Question 4

Prove the trigonometric identity: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$. Show all steps clearly.

Accepted Answer

To prove: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$ 
Take the Left Hand Side (LHS): $	ext{LHS} = \frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta}$ 
Factor out $\sin 	heta$ from the numerator and $\cos 	heta$ from the denominator: $	e

Question 5

Prove the identity: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + 	an A$. Show all steps clearly.

Accepted Answer

To prove: $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + 	an A$ 
Take the Left Hand Side (LHS): $	ext{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$ 
Rationalize the denominator inside the square root by multiplying both numerator and denominator by $(1 + \sin A)$: $	ext{LHS} = \sqrt{\frac{(1 +

Introduction to Trigonometry