OP Malhotra Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Exercise 4 (A)

Prove That

Question 1. \( 1 - \cos^2 \theta = \sin^2 \theta \).
Answer:
L.H.S. \( = 1 - \cos^2 \theta \)
We know that \( \cos \theta = \frac{x}{r} \), so \( \cos^2 \theta = \frac{x^2}{r^2} \).
\( \implies = 1 - \frac{x^2}{r^2} \)
\( \implies = \frac{r^2 - x^2}{r^2} \)
Since \( x^2 + y^2 = r^2 \) (Pythagorean theorem), we have \( r^2 - x^2 = y^2 \).
\( \implies = \frac{y^2}{r^2} \)
We also know that \( \sin \theta = \frac{y}{r} \), so \( \sin^2 \theta = \frac{y^2}{r^2} \).
\( \implies = \sin^2 \theta \)
\( = R.H.S. \)
In simple words: We start with the left side of the equation. By replacing `cos²θ` with its fraction form and using the Pythagorean theorem, we can change the expression to `y²/r²`, which is the same as `sin²θ`. This shows both sides are equal.

🎯 Exam Tip: Remember the fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity is derived directly from the Pythagorean theorem in a right-angled triangle.

Question 2. \( \sqrt{1-\sin^2 \theta} = \cos \theta \)
Answer:
L.H.S. \( = \sqrt{1-\sin^2 \theta} \)
We know the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). From this, \( 1 - \sin^2 \theta = \cos^2 \theta \).
\( \implies = \sqrt{\cos^2 \theta} \)
The square root of `cos²θ` is the absolute value of `cos θ`, written as \( |\cos \theta| \).
\( \implies = |\cos \theta| \)
For the given identity to be true as \( \cos \theta \) (without the absolute value), we assume that \( \cos \theta \geq 0 \).
\( \implies = \cos \theta \)
\( = R.H.S. \)
In simple words: This problem uses the basic rule `sin²θ + cos²θ = 1`. If we rearrange it, `1 - sin²θ` becomes `cos²θ`. Then, the square root of `cos²θ` is `cos θ`, assuming `cos θ` is not negative.

🎯 Exam Tip: When taking the square root of a squared trigonometric function, always remember to consider the absolute value, as \( \sqrt{x^2} = |x| \). The context of the problem often implies the positive root.

Question 3. \( \sec \alpha \sqrt{1-\sin^2 \alpha} = 1 \)
Answer:
L.H.S. \( = \sec \alpha \sqrt{1-\sin^2 \alpha} \)
Using the identity \( 1 - \sin^2 \alpha = \cos^2 \alpha \):
\( \implies = \sec \alpha \sqrt{\cos^2 \alpha} \)
\( \implies = \sec \alpha |\cos \alpha| \)
We know that \( \sec \alpha = \frac{1}{\cos \alpha} \). For the expression to simplify to 1, we must have \( \cos \alpha > 0 \). This happens when \( \alpha \) lies in the first or fourth quadrant.
\( \implies = \frac{1}{\cos \alpha} \cdot \cos \alpha \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: We start with the left side. First, we change `1 - sin²α` into `cos²α`. Then we take the square root, which gives us the absolute value of `cos α`. Because `sec α` and `cos α` are opposite (reciprocal) functions, they multiply to 1, as long as `cos α` is positive.

🎯 Exam Tip: Pay close attention to the conditions under which an identity holds true, especially when dealing with square roots and absolute values. Here, `cos α > 0` is crucial.

Question 4. \( \frac{\sec^2 \theta-1}{\tan^2 \theta} = 1 \)
Answer:
L.H.S. \( = \frac{\sec^2 \theta-1}{\tan^2 \theta} \)
We know that \( \sec \theta = \frac{r}{x} \) and \( \tan \theta = \frac{y}{x} \).
\( \implies = \frac{\frac{r^2}{x^2}-1}{\frac{y^2}{x^2}} \)
\( \implies = \frac{\frac{r^2-x^2}{x^2}}{\frac{y^2}{x^2}} \)
We can simplify the fraction by canceling out the common denominator \( x^2 \).
\( \implies = \frac{r^2-x^2}{y^2} \)
From the Pythagorean theorem, \( x^2+y^2=r^2 \), so \( r^2-x^2=y^2 \).
\( \implies = \frac{y^2}{y^2} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: This problem uses another main trigonometry rule: `1 + tan²θ = sec²θ`. This means `sec²θ - 1` is exactly equal to `tan²θ`. So, when you divide `tan²θ` by itself, the answer is always 1.

🎯 Exam Tip: Remember the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), which means \( \sec^2 \theta - 1 = \tan^2 \theta \). This direct substitution is the quickest way to solve this type of problem.

Question 5. \( \sec^2 \theta - 1 - \tan^2 \theta = 0 \).
Answer:
L.H.S. \( = \sec^2 \theta - 1 - \tan^2 \theta \)
We know that \( \sec \theta = \frac{r}{x} \) and \( \tan \theta = \frac{y}{x} \).
\( \implies = \frac{r^2}{x^2}-1-\frac{y^2}{x^2} \)
We combine these terms over the common denominator \( x^2 \).
\( \implies = \frac{r^2-x^2-y^2}{x^2} \)
From the Pythagorean theorem, \( x^2+y^2=r^2 \), so \( r^2 = x^2+y^2 \).
\( \implies = \frac{r^2-(x^2+y^2)}{x^2} \)
\( \implies = \frac{r^2-r^2}{x^2} \)
\( \implies = \frac{0}{x^2} \)
\( \implies = 0 \)
\( = R.H.S. \)
In simple words: We know the rule `sec²θ = 1 + tan²θ`. If we move `1` and `tan²θ` to the other side, the equation becomes `sec²θ - 1 - tan²θ = 0`. This identity is a simple rearrangement of the basic one.

🎯 Exam Tip: Rearranging fundamental identities like \( 1 + \tan^2 \theta = \sec^2 \theta \) is a common technique to prove other identities. Practice recognizing these relationships quickly.

Question 6. \( \frac{\sin^2 \theta+\cos^2 \theta}{\sec^2 \theta-\tan^2 \theta} = 1 \).
Answer:
L.H.S. \( = \frac{\sin^2 \theta+\cos^2 \theta}{\sec^2 \theta-\tan^2 \theta} \)
We use the definitions \( \sin \theta = \frac{y}{r} \), \( \cos \theta = \frac{x}{r} \), \( \sec \theta = \frac{r}{x} \), and \( \tan \theta = \frac{y}{x} \).
\( \implies = \frac{\frac{y^2}{r^2}+\frac{x^2}{r^2}}{\frac{r^2}{x^2}-\frac{y^2}{x^2}} \)
\( \implies = \frac{\frac{y^2+x^2}{r^2}}{\frac{r^2-y^2}{x^2}} \)
Using the Pythagorean identity \( x^2+y^2=r^2 \), we can replace the numerators and denominators.
\( \implies = \frac{\frac{r^2}{r^2}}{\frac{x^2}{x^2}} \)
\( \implies = \frac{1}{1} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: The top part `sin²θ + cos²θ` always equals 1. The bottom part `sec²θ - tan²θ` also always equals 1. So, when you divide 1 by 1, the answer is 1.

🎯 Exam Tip: This problem combines two key identities: \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \sec^2 \theta - \tan^2 \theta = 1 \). Recognizing these simplifies the expression immediately.

Question 7. \( 1 - \cos^2 \theta - \sin^2 \theta = 0 \).
Answer:
L.H.S. \( = 1 - \cos^2 \theta - \sin^2 \theta \)
We can factor out a negative sign from the last two terms.
\( \implies = 1 - (\cos^2 \theta + \sin^2 \theta) \)
Using the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \implies = 1 - 1 \)
\( \implies = 0 \)
\( = R.H.S. \)
In simple words: This problem asks us to show that `1 - cos²θ - sin²θ` is zero. Since `cos²θ + sin²θ` is always 1, we can replace that part with 1. Then `1 - 1` is 0.

🎯 Exam Tip: When you see `sin²θ` and `cos²θ` together, always think of the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). It's a very common simplification.

Question 8. \( \sin^4 \theta + \sin^2 \theta \cos^2 \theta = \sin^2 \theta \).
Answer:
L.H.S. \( = \sin^4 \theta + \sin^2 \theta \cos^2 \theta \)
We can take out a common factor of \( \sin^2 \theta \).
\( \implies = \sin^2 \theta (\sin^2 \theta + \cos^2 \theta) \)
Using the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \implies = \sin^2 \theta (1) \)
\( \implies = \sin^2 \theta \)
\( = R.H.S. \)

Aliter:
L.H.S. \( = \sin^4 \theta + \sin^2 \theta \cos^2 \theta \)
Using definitions \( \sin \theta = \frac{y}{r} \) and \( \cos \theta = \frac{x}{r} \).
\( \implies = \frac{y^4}{r^4} + \frac{y^2}{r^2} \cdot \frac{x^2}{r^2} \)
\( \implies = \frac{y^4}{r^4} + \frac{y^2 x^2}{r^4} \)
\( \implies = \frac{y^2(y^2+x^2)}{r^4} \)
From the Pythagorean identity \( x^2+y^2=r^2 \).
\( \implies = \frac{y^2 r^2}{r^4} \)
\( \implies = \frac{y^2}{r^2} \)
\( \implies = \sin^2 \theta \)
\( = R.H.S. \)
In simple words: To solve this, we can take out `sin²θ` as a common part from both terms on the left side. What's left inside the bracket is `sin²θ + cos²θ`, which is equal to 1. So, the whole thing simplifies to `sin²θ`.

🎯 Exam Tip: Look for common factors first. Factoring out `sin²θ` immediately reveals the Pythagorean identity, leading to a quick solution.

Question 9. \( \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta = 1 \).
Answer:
L.H.S. \( = \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta \)
This expression looks like the expansion of a perfect square \( (a+b)^2 = a^2+2ab+b^2 \). Here, \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \).
\( \implies = (\sin^2 \theta + \cos^2 \theta)^2 \)
Using the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \implies = (1)^2 \)
\( \implies = 1 \)
\( = R.H.S. \)

Aliter:
L.H.S. \( = \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta \)
Using definitions \( \sin \theta = \frac{y}{r} \) and \( \cos \theta = \frac{x}{r} \).
\( \implies = \frac{y^4}{r^4} + 2 \frac{y^2}{r^2} \cdot \frac{x^2}{r^2} + \frac{x^4}{r^4} \)
\( \implies = \frac{y^4 + 2y^2 x^2 + x^4}{r^4} \)
The numerator is a perfect square \( (y^2+x^2)^2 \).
\( \implies = \frac{(y^2+x^2)^2}{r^4} \)
From the Pythagorean identity \( x^2+y^2=r^2 \).
\( \implies = \frac{(r^2)^2}{r^4} \)
\( \implies = \frac{r^4}{r^4} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: The left side of the equation is a special type of sum called a perfect square. It's like `(a + b)²` where `a` is `sin²θ` and `b` is `cos²θ`. Since `sin²θ + cos²θ` equals 1, squaring 1 still gives 1.

🎯 Exam Tip: Recognize algebraic patterns like perfect squares. \( \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta \) is exactly \( (\sin^2 \theta + \cos^2 \theta)^2 \), which simplifies using the fundamental identity.

Question 10. \( \cos^2 \theta (\operatorname{cosec}^2 \theta - \cot^2 \theta) = \cot^2 \theta \).
Answer:
L.H.S. \( = \cos^2 \theta (\operatorname{cosec}^2 \theta - \cot^2 \theta) \)
We use the identity \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \), which means \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \).
\( \implies = \cos^2 \theta (1) \)
\( \implies = \cos^2 \theta \)

Aliter:
L.H.S. \( = \cos^2 \theta (\operatorname{cosec}^2 \theta - \cot^2 \theta) \)
Using definitions \( \cos \theta = \frac{x}{r} \), \( \operatorname{cosec} \theta = \frac{r}{y} \), and \( \cot \theta = \frac{x}{y} \).
\( \implies = \frac{x^2}{r^2}\left(\frac{r^2}{y^2}-\frac{x^2}{y^2}\right) \)
\( \implies = \frac{x^2}{r^2}\left(\frac{r^2-x^2}{y^2}\right) \)
From the Pythagorean identity \( x^2+y^2=r^2 \), so \( r^2-x^2=y^2 \).
\( \implies = \frac{x^2}{r^2}\left(\frac{y^2}{y^2}\right) \)
\( \implies = \frac{x^2}{r^2}(1) \)
\( \implies = \frac{x^2}{r^2} \)
\( \implies = \cos^2 \theta \)
In simple words: We look at the left side. The part inside the bracket, `cosec²θ - cot²θ`, is a known identity that equals 1. So, the whole expression becomes `cos²θ` multiplied by 1, which is just `cos²θ`.

🎯 Exam Tip: There might be a typo in the question's RHS, which states \( \cot^2 \theta \). Based on the identity \( \operatorname{cosec}^2 \theta - \cot^2 \theta = 1 \), the LHS simplifies to \( \cos^2 \theta \). Always present the solution that follows logically from the given steps and identities.

Question 11. \( \frac{\sin \theta \operatorname{cosec} \theta \tan \theta \cot \theta}{\sin^2 \theta+\cos^2 \theta} = 1 \).
Answer:
L.H.S. \( = \frac{\sin \theta \operatorname{cosec} \theta \tan \theta \cot \theta}{\sin^2 \theta+\cos^2 \theta} \)
We use the reciprocal identities \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \) and \( \cot \theta = \frac{1}{\tan \theta} \). Also, \( \sin^2 \theta + \cos^2 \theta = 1 \).
\( \implies = \frac{\sin \theta \cdot \frac{1}{\sin \theta} \cdot \tan \theta \cdot \frac{1}{\tan \theta}}{1} \)
\( \implies = \frac{1 \cdot 1}{1} \)
\( \implies = \frac{1}{1} \)
\( \implies = 1 \)
\( = R.H.S. \)

Aliter:
L.H.S. \( = \frac{\sin \theta \operatorname{cosec} \theta \tan \theta \cot \theta}{\sin^2 \theta+\cos^2 \theta} \)
Using definitions \( \sin \theta = \frac{y}{r} \), \( \operatorname{cosec} \theta = \frac{r}{y} \), \( \tan \theta = \frac{y}{x} \), and \( \cot \theta = \frac{x}{y} \).
\( \implies = \frac{\frac{y}{r} \cdot \frac{r}{y} \cdot \frac{y}{x} \cdot \frac{x}{y}}{\frac{y^2}{r^2}+\frac{x^2}{r^2}} \)
\( \implies = \frac{1 \cdot 1}{\frac{y^2+x^2}{r^2}} \)
From the Pythagorean identity \( x^2+y^2=r^2 \).
\( \implies = \frac{1}{\frac{r^2}{r^2}} \)
\( \implies = \frac{1}{1} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: This problem is simpler than it looks because of helpful identities. `sin θ` and `cosec θ` cancel each other out to 1. `tan θ` and `cot θ` also cancel out to 1. And `sin²θ + cos²θ` in the bottom is also 1. So, everything simplifies to `1/1`, which is 1.

🎯 Exam Tip: When proving identities, always simplify terms using reciprocal identities (\( \sin \theta \operatorname{cosec} \theta = 1 \), \( \tan \theta \cot \theta = 1 \)) and Pythagorean identities (\( \sin^2 \theta + \cos^2 \theta = 1 \)) first.

Question 12. \( \frac{\sin^2 30^{\circ}+\cos^2 30^{\circ}}{\sec^2 57^{\circ}-\tan^2 57^{\circ}} = 1 \)
Answer:
L.H.S. \( = \frac{\sin^2 30^{\circ}+\cos^2 30^{\circ}}{\sec^2 57^{\circ}-\tan^2 57^{\circ}} \)
We use the fundamental identity \( \sin^2 \theta + \cos^2 \theta = 1 \) for the numerator (where \( \theta = 30^{\circ} \)).
And we use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \) for the denominator (where \( \theta = 57^{\circ} \)).
\( \implies = \frac{1}{1} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: The top part `sin²30° + cos²30°` is always 1, no matter the angle. The bottom part `sec²57° - tan²57°` is also always 1, based on another rule. So, dividing 1 by 1 gives 1.

🎯 Exam Tip: This question tests your knowledge of fundamental trigonometric identities. The specific angle values (\( 30^{\circ} \) and \( 57^{\circ} \)) are distractions; the identities hold for any valid angle.

Question 13. \( \cos A \tan A = \sin A \).
Answer:
L.H.S. \( = \cos A \tan A \)
We use the quotient identity \( \tan A = \frac{\sin A}{\cos A} \).
\( \implies = \cos A \cdot \frac{\sin A}{\cos A} \)
The \( \cos A \) terms cancel out, as long as \( \cos A \neq 0 \).
\( \implies = \sin A \)
\( = R.H.S. \)
In simple words: We start with the left side. We know that `tan A` can be written as `sin A` divided by `cos A`. When we put this into the equation, the `cos A` parts cancel each other out, leaving only `sin A`.

🎯 Exam Tip: When you see `tan A` or `cot A` in an expression, consider rewriting them in terms of `sin A` and `cos A` using quotient identities (\( \tan A = \frac{\sin A}{\cos A} \), \( \cot A = \frac{\cos A}{\sin A} \)).

Question 14. \( \sin^4 A \operatorname{cosec}^2 A + \cos^4 A \sec^2 A = 1 \).
Answer:
L.H.S. \( = \sin^4 A \operatorname{cosec}^2 A + \cos^4 A \sec^2 A \)
We use the reciprocal identities \( \operatorname{cosec}^2 A = \frac{1}{\sin^2 A} \) and \( \sec^2 A = \frac{1}{\cos^2 A} \).
\( \implies = \sin^4 A \cdot \frac{1}{\sin^2 A} + \cos^4 A \cdot \frac{1}{\cos^2 A} \)
We can cancel out the squared terms.
\( \implies = \sin^2 A + \cos^2 A \)
Using the fundamental identity \( \sin^2 A + \cos^2 A = 1 \).
\( \implies = 1 \)
\( = R.H.S. \)

Aliter:
L.H.S. \( = \sin^4 A \operatorname{cosec}^2 A + \cos^4 A \sec^2 A \)
Using definitions \( \sin A = \frac{y}{r} \), \( \cos A = \frac{x}{r} \), \( \operatorname{cosec} A = \frac{r}{y} \), and \( \sec A = \frac{r}{x} \).
\( \implies = \frac{y^4}{r^4} \cdot \frac{r^2}{y^2} + \frac{x^4}{r^4} \cdot \frac{r^2}{x^2} \)
\( \implies = \frac{y^2}{r^2} + \frac{x^2}{r^2} \)
\( \implies = \frac{y^2+x^2}{r^2} \)
From the Pythagorean identity \( x^2+y^2=r^2 \).
\( \implies = \frac{r^2}{r^2} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: For the left side, we change `cosec²A` to `1/sin²A` and `sec²A` to `1/cos²A`. This lets us cancel out `sin²A` and `cos²A` from the `sin⁴A` and `cos⁴A` terms. What's left is `sin²A + cos²A`, which is always equal to 1.

🎯 Exam Tip: When dealing with higher powers, try to reduce them using reciprocal identities. This often leads to simpler expressions that can be solved with fundamental identities.

Question 15. \( \sin^2 A \cot^2 A + \cos^2 A \tan^2 A = 1 \).
Answer:
L.H.S. \( = \sin^2 A \cot^2 A + \cos^2 A \tan^2 A \)
We use the quotient identities \( \cot A = \frac{\cos A}{\sin A} \) and \( \tan A = \frac{\sin A}{\cos A} \).
\( \implies = \sin^2 A \cdot \frac{\cos^2 A}{\sin^2 A} + \cos^2 A \cdot \frac{\sin^2 A}{\cos^2 A} \)
The \( \sin^2 A \) and \( \cos^2 A \) terms cancel out.
\( \implies = \cos^2 A + \sin^2 A \)
Using the fundamental identity \( \sin^2 A + \cos^2 A = 1 \).
\( \implies = 1 \)
\( = R.H.S. \)

Aliter:
L.H.S. \( = \sin^2 A \cot^2 A + \cos^2 A \tan^2 A \)
Using definitions \( \sin A = \frac{y}{r} \), \( \cos A = \frac{x}{r} \), \( \tan A = \frac{y}{x} \), and \( \cot A = \frac{x}{y} \).
\( \implies = \frac{y^2}{r^2} \cdot \frac{x^2}{y^2} + \frac{x^2}{r^2} \cdot \frac{y^2}{x^2} \)
\( \implies = \frac{x^2}{r^2} + \frac{y^2}{r^2} \)
\( \implies = \frac{x^2+y^2}{r^2} \)
From the Pythagorean identity \( x^2+y^2=r^2 \).
\( \implies = \frac{r^2}{r^2} \)
\( \implies = 1 \)
\( = R.H.S. \)
In simple words: For the left side, we replace `cot²A` with `cos²A/sin²A` and `tan²A` with `sin²A/cos²A`. This makes some parts cancel out, leaving `cos²A + sin²A`. This sum is always equal to 1.

🎯 Exam Tip: When dealing with `tan` and `cot`, rewriting them in terms of `sin` and `cos` is often the most effective first step towards simplifying the expression and proving identities.