RBSE Solutions Class 9 Maths Chapter 14 Trigonometric Ratios of Acute Angles Important Questions

Multiple Choice Questions

 

Question 1. The value of \( \text{sec}^2 50^\circ - \text{tan}^2 50^\circ \) is equal to:
(A) -1
(B) 0
(C) 1
(D) 2
Answer: (C) 1
In simple words: This is a basic trigonometric identity. For any angle \( \theta \), the value of \( \text{sec}^2 \theta - \text{tan}^2 \theta \) is always 1.

🎯 Exam Tip: Remember the fundamental trigonometric identities: \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \), \( 1 + \text{tan}^2 \theta = \text{sec}^2 \theta \), and \( 1 + \text{cot}^2 \theta = \text{cosec}^2 \theta \).

 

Question 2. \( \frac { 1 }{ \text{cosec } \theta - \text{cot } \theta } \) is equal to:
(A) \( \text{cosec } \theta + \text{cot } \theta \)
(B) \( \frac { 1 }{ \text{cosec } \theta + \text{cot } \theta } \)
(C) \( \text{cosec } \theta - \text{cot } \theta \)
(D) \( \text{tan } \theta \)
Answer: (A) \( \text{cosec } \theta + \text{cot } \theta \)
In simple words: To simplify this fraction, we multiply the top and bottom by \( \text{cosec } \theta + \text{cot } \theta \). This uses the difference of squares identity, making the denominator 1 because \( \text{cosec}^2 \theta - \text{cot}^2 \theta = 1 \).

🎯 Exam Tip: When you see expressions like \( \frac{1}{A-B} \) with trigonometric terms, try multiplying by the conjugate \( \frac{A+B}{A+B} \) to simplify, especially if \( A^2 - B^2 \) is a known identity.

 

Question 3. \( \frac { \text{tan } \phi }{ \sqrt { 1 + \text{tan}^2 \phi } } \) is equal to:
(A) \( \frac { \text{tan } \phi }{ 1 + \text{tan } \phi } \)
(B) \( \frac { \text{tan } \phi }{ \text{sec}^2 \phi } \)
(C) \( \text{sin } \phi \)
(D) \( \text{cos } \phi \)
Answer: (C) \( \text{sin } \phi \)
In simple words: We know that \( 1 + \text{tan}^2 \phi = \text{sec}^2 \phi \). So, the square root becomes \( \text{sec } \phi \). Then, we can change \( \text{tan } \phi \) to \( \frac{\text{sin } \phi}{\text{cos } \phi} \) and \( \text{sec } \phi \) to \( \frac{1}{\text{cos } \phi} \). When we divide, the \( \text{cos } \phi \) terms cancel out, leaving just \( \text{sin } \phi \).

🎯 Exam Tip: Convert all tangent and secant terms to sine and cosine when simplifying complex trigonometric expressions. This often makes the solution clearer and easier to find.

 

Question 5. If \( 3 \text{ cot } \phi = 2 \) then the value of \( \text{cosec}^2 \phi \) is equal to:
(A) \( \frac { \sqrt{3} }{ 3 } \)
(B) \( \frac { 13 }{ 9 } \)
(C) \( \frac { 9 }{ 13 } \)
(D) \( \frac { \sqrt{3} }{ 3 } \)
Answer: (B) \( \frac { 13 }{ 9 } \)
In simple words: From \( 3 \text{ cot } \phi = 2 \), we find that \( \text{cot } \phi = \frac{2}{3} \). Then, we use the identity \( \text{cosec}^2 \phi = 1 + \text{cot}^2 \phi \). We substitute the value of \( \text{cot } \phi \) into this identity to get the answer.

🎯 Exam Tip: Always recall the basic Pythagorean identities in trigonometry when one ratio is given and another needs to be found. They provide a direct path to the solution.

Very Short Answer Type Questions

 

Question 1. Write the value of \( \text{cosec}^2 50^\circ - \text{cot}^2 50^\circ \).
Answer: \( \text{cosec}^2 \theta - \text{cot}^2 \theta = 1 \). So, \( \text{cosec}^2 50^\circ - \text{cot}^2 50^\circ = 1 \). This identity is true for any valid angle \( \theta \), which includes \( 50^\circ \).
In simple words: The value is 1. This is a basic rule in trigonometry that never changes, no matter what angle you use.

🎯 Exam Tip: Always be mindful of the angle given; while this identity is angle-independent, other formulas might not be, so always read the question carefully.

 

Question 2. Prove that \( \frac { \text{sin } \theta }{ \text{cosec } \theta } + \frac { \text{cos } \theta }{ \text{sec } \theta } = 1 \).
Answer:
L.H.S. \( = \frac { \text{sin } \theta }{ \text{cosec } \theta } + \frac { \text{cos } \theta }{ \text{sec } \theta } \)
\( = \text{sin } \theta \cdot \text{sin } \theta + \text{cos } \theta \cdot \text{cos } \theta \) (Since \( \text{cosec } \theta = \frac{1}{\text{sin } \theta} \) and \( \text{sec } \theta = \frac{1}{\text{cos } \theta} \))
\( = \text{sin}^2 \theta + \text{cos}^2 \theta \)
\( = 1 \)
\( = \text{R.H.S.} \) Hence proved. This identity shows the relationship between reciprocal trigonometric functions.
In simple words: We can change \( \text{cosec } \theta \) to \( \frac{1}{\text{sin } \theta} \) and \( \text{sec } \theta \) to \( \frac{1}{\text{cos } \theta} \). Then, the fractions become \( \text{sin}^2 \theta \) and \( \text{cos}^2 \theta \). Adding them gives 1, which proves the statement.

🎯 Exam Tip: When proving identities, always start with the more complex side (usually the L.H.S.) and simplify it step-by-step until it matches the R.H.S.

 

Question 3. Prove that \( \text{sin } \theta \text{ cosec } \theta + \text{cos } \theta \text{ sec } \theta = 2 \).
Answer:
L.H.S. \( = \text{sin } \theta \text{ cosec } \theta + \text{cos } \theta \text{ sec } \theta \)
\( = \text{sin } \theta \cdot \frac{1}{\text{sin } \theta} + \text{cos } \theta \cdot \frac{1}{\text{cos } \theta} \) (Using reciprocal identities for cosec and sec)
\( = 1 + 1 \)
\( = 2 \)
\( = \text{R.H.S.} \) Hence proved. This identity highlights how reciprocal functions cancel each other out.
In simple words: We know that \( \text{cosec } \theta \) is \( \frac{1}{\text{sin } \theta} \) and \( \text{sec } \theta \) is \( \frac{1}{\text{cos } \theta} \). So, \( \text{sin } \theta \) times \( \text{cosec } \theta \) becomes 1, and \( \text{cos } \theta \) times \( \text{sec } \theta \) becomes 1. Adding 1 and 1 gives 2, which proves the statement.

🎯 Exam Tip: Always remember the reciprocal relations: \( \text{sin } \theta = \frac{1}{\text{cosec } \theta} \), \( \text{cos } \theta = \frac{1}{\text{sec } \theta} \), and \( \text{tan } \theta = \frac{1}{\text{cot } \theta} \).

 

Question 4. Prove that \( \text{cot } \phi = \sqrt { \frac { 1 - \text{sin}^2 \phi }{ \text{sin}^2 \phi } } \).
Answer:
R.H.S. \( = \sqrt { \frac { 1 - \text{sin}^2 \phi }{ \text{sin}^2 \phi } } \)
\( = \sqrt { \frac { \text{cos}^2 \phi }{ \text{sin}^2 \phi } } \) (Since \( 1 - \text{sin}^2 \phi = \text{cos}^2 \phi \))
\( = \frac { \sqrt { \text{cos}^2 \phi } }{ \sqrt { \text{sin}^2 \phi } } \)
\( = \frac { \text{cos } \phi }{ \text{sin } \phi } \)
\( = \text{cot } \phi \)
\( = \text{L.H.S.} \) Hence proved. This shows how cotangent can be expressed using sine and cosine functions.
In simple words: We know that \( 1 - \text{sin}^2 \phi \) is the same as \( \text{cos}^2 \phi \). So, the fraction inside the square root becomes \( \frac{\text{cos}^2 \phi}{\text{sin}^2 \phi} \). Taking the square root of both top and bottom gives \( \frac{\text{cos } \phi}{\text{sin } \phi} \), which is equal to \( \text{cot } \phi \).

🎯 Exam Tip: When dealing with square roots in trigonometric identities, try to simplify the expression inside the root first using Pythagorean identities before taking the square root.

 

Question 5. Write the value of \( (1 - \text{cos } \theta)(1 + \text{cos } \theta)(1 + \text{cot}^2 \theta) \).
Answer:
Expression \( = (1 - \text{cos } \theta)(1 + \text{cos } \theta)(1 + \text{cot}^2 \theta) \)
\( = (1 - \text{cos}^2 \theta)(\text{cosec}^2 \theta) \) (Using \( (A-B)(A+B) = A^2-B^2 \) and \( 1 + \text{cot}^2 \theta = \text{cosec}^2 \theta \))
\( = \text{sin}^2 \theta \cdot \text{cosec}^2 \theta \) (Since \( 1 - \text{cos}^2 \theta = \text{sin}^2 \theta \))
\( = \text{sin}^2 \theta \cdot \frac{1}{\text{sin}^2 \theta} \) (Since \( \text{cosec } \theta = \frac{1}{\text{sin } \theta} \))
\( = 1 \)
In simple words: First, \( (1 - \text{cos } \theta)(1 + \text{cos } \theta) \) becomes \( 1 - \text{cos}^2 \theta \), which is \( \text{sin}^2 \theta \). Also, \( (1 + \text{cot}^2 \theta) \) is \( \text{cosec}^2 \theta \). When we multiply \( \text{sin}^2 \theta \) by \( \text{cosec}^2 \theta \), it simplifies to 1 because they are reciprocals.

🎯 Exam Tip: Look for opportunities to apply both algebraic identities (like difference of squares) and trigonometric identities simultaneously to simplify expressions efficiently.

 

Question 6. If \( \text{cos}^2 \theta - \text{sin}^2 \theta = \frac{1}{2} \), find the value of \( \text{cos}^4 \theta - \text{sin}^4 \theta \).
Answer:
Expression \( = \text{cos}^4 \theta - \text{sin}^4 \theta \)
\( = (\text{cos}^2 \theta)^2 - (\text{sin}^2 \theta)^2 \)
\( = (\text{cos}^2 \theta - \text{sin}^2 \theta)(\text{cos}^2 \theta + \text{sin}^2 \theta) \) (Using \( A^2 - B^2 = (A-B)(A+B) \))
\( = (\frac{1}{2})(1) \) (Given \( \text{cos}^2 \theta - \text{sin}^2 \theta = \frac{1}{2} \) and \( \text{cos}^2 \theta + \text{sin}^2 \theta = 1 \))
\( = \frac{1}{2} \)
In simple words: We can rewrite \( \text{cos}^4 \theta - \text{sin}^4 \theta \) using the difference of squares rule. This gives us \( (\text{cos}^2 \theta - \text{sin}^2 \theta) \) multiplied by \( (\text{cos}^2 \theta + \text{sin}^2 \theta) \). We are given the first part is \( \frac{1}{2} \), and the second part is always 1. So, the answer is simply \( \frac{1}{2} \).

🎯 Exam Tip: Recognize higher powers of trigonometric functions as opportunities to use algebraic factorization (like difference of squares or cubes) combined with basic identities.

 

Question 7. Prove that \( \text{tan}^2 A + \text{tan}^4 A = \text{sec}^4 A - \text{sec}^2 A \).
Answer:
L.H.S. \( = \text{tan}^2 A + \text{tan}^4 A \)
\( = \text{tan}^2 A (1 + \text{tan}^2 A) \) (Factor out \( \text{tan}^2 A \))
\( = (\text{sec}^2 A - 1)(\text{sec}^2 A) \) (Since \( 1 + \text{tan}^2 A = \text{sec}^2 A \), and \( \text{tan}^2 A = \text{sec}^2 A - 1 \))
\( = \text{sec}^4 A - \text{sec}^2 A \)
\( = \text{R.H.S.} \) Hence proved. This identity shows the flexibility in expressing tangent in terms of secant.
In simple words: Start with the left side and take out \( \text{tan}^2 A \) as a common factor. Then, use the rule that \( 1 + \text{tan}^2 A = \text{sec}^2 A \). Also, change the remaining \( \text{tan}^2 A \) to \( \text{sec}^2 A - 1 \). Multiply these terms out to get the right side of the equation.

🎯 Exam Tip: When proving identities, if one side contains a higher power of a trigonometric function, consider factoring it out or using identities to express it in terms of other functions.

 

Question 8. Find the value of \( \frac { 1 }{ \text{sec}^2 \theta } + \frac { 1 }{ \text{cosec}^2 \theta } \).
Answer:
Expression \( = \frac { 1 }{ \text{sec}^2 \theta } + \frac { 1 }{ \text{cosec}^2 \theta } \)
\( = \text{cos}^2 \theta + \text{sin}^2 \theta \) (Since \( \frac{1}{\text{sec } \theta} = \text{cos } \theta \) and \( \frac{1}{\text{cosec } \theta} = \text{sin } \theta \))
\( = 1 \)
In simple words: We know that \( \frac{1}{\text{sec}^2 \theta} \) is the same as \( \text{cos}^2 \theta \), and \( \frac{1}{\text{cosec}^2 \theta} \) is the same as \( \text{sin}^2 \theta \). When we add \( \text{cos}^2 \theta \) and \( \text{sin}^2 \theta \), we always get 1.

🎯 Exam Tip: Always remember the reciprocal identities and the fundamental Pythagorean identity \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \). They are very useful for simplifying expressions quickly.

 

Question 9. Find the value of \( (\text{cos}^2 \theta - 1)(1 + \text{cot}^2 \theta) \).
Answer:
Expression \( = (\text{cos}^2 \theta - 1)(1 + \text{cot}^2 \theta) \)
\( = - (1 - \text{cos}^2 \theta)(\text{cosec}^2 \theta) \) (Factor out -1 from the first term and use \( 1 + \text{cot}^2 \theta = \text{cosec}^2 \theta \))
\( = - (\text{sin}^2 \theta)(\text{cosec}^2 \theta) \) (Since \( 1 - \text{cos}^2 \theta = \text{sin}^2 \theta \))
\( = - (\text{sin}^2 \theta)(\frac{1}{\text{sin}^2 \theta}) \) (Since \( \text{cosec } \theta = \frac{1}{\text{sin } \theta} \))
\( = -1 \)
In simple words: First, \( (\text{cos}^2 \theta - 1) \) is the same as \( - \text{sin}^2 \theta \). Then, \( (1 + \text{cot}^2 \theta) \) is \( \text{cosec}^2 \theta \). When we multiply \( - \text{sin}^2 \theta \) by \( \text{cosec}^2 \theta \), it simplifies to -1 because \( \text{sin}^2 \theta \) and \( \text{cosec}^2 \theta \) are reciprocals, and there's a negative sign.

🎯 Exam Tip: Pay close attention to signs, especially when rearranging terms like \( (\text{cos}^2 \theta - 1) \) to \( - (1 - \text{cos}^2 \theta) \).

 

Question 10. Evaluate \( (1 - \text{sin}^2 \theta)(1 + \text{tan}^2 \theta) \).
Answer:
Expression \( = (1 - \text{sin}^2 \theta)(1 + \text{tan}^2 \theta) \)
\( = (\text{cos}^2 \theta)(\text{sec}^2 \theta) \) (Since \( 1 - \text{sin}^2 \theta = \text{cos}^2 \theta \) and \( 1 + \text{tan}^2 \theta = \text{sec}^2 \theta \))
\( = \text{cos}^2 \theta \cdot \frac{1}{\text{cos}^2 \theta} \) (Since \( \text{sec } \theta = \frac{1}{\text{cos } \theta} \))
\( = 1 \)
In simple words: We know that \( (1 - \text{sin}^2 \theta) \) is \( \text{cos}^2 \theta \) and \( (1 + \text{tan}^2 \theta) \) is \( \text{sec}^2 \theta \). Since \( \text{sec}^2 \theta \) is the reciprocal of \( \text{cos}^2 \theta \), multiplying them together gives 1.

🎯 Exam Tip: Always remember the fundamental identities \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \) and \( 1 + \text{tan}^2 \theta = \text{sec}^2 \theta \). They are keys to simplifying many expressions.

Short Answer Type Questions

 

Question 1. If \( \text{sec } \theta = \frac{13}{12} \), then find the value of \( \frac { 1 - \text{tan } \theta }{ 1 + \text{tan } \theta } \).
Answer:
Given \( \text{sec } \theta = \frac{13}{12} \)
We know that \( \text{tan}^2 \theta = \text{sec}^2 \theta - 1 \)
\( \implies \text{tan}^2 \theta = (\frac{13}{12})^2 - 1 \)
\( \implies \text{tan}^2 \theta = \frac{169}{144} - 1 \)
\( \implies \text{tan}^2 \theta = \frac{169 - 144}{144} \)
\( \implies \text{tan}^2 \theta = \frac{25}{144} \)
\( \implies \text{tan } \theta = \sqrt{\frac{25}{144}} \)
\( \implies \text{tan } \theta = \frac{5}{12} \)
Now, we need to find the value of \( \frac { 1 - \text{tan } \theta }{ 1 + \text{tan } \theta } \)
\( = \frac { 1 - \frac{5}{12} }{ 1 + \frac{5}{12} } \)
\( = \frac { \frac{12 - 5}{12} }{ \frac{12 + 5}{12} } \)
\( = \frac { \frac{7}{12} }{ \frac{17}{12} } \)
\( = \frac{7}{12} \cdot \frac{12}{17} \)
\( = \frac{7}{17} \)
In simple words: First, use the identity \( \text{tan}^2 \theta = \text{sec}^2 \theta - 1 \) to find the value of \( \text{tan } \theta \). Since \( \text{sec } \theta = \frac{13}{12} \), \( \text{tan } \theta \) comes out to be \( \frac{5}{12} \). Then, put this value of \( \text{tan } \theta \) into the given fraction \( \frac{1 - \text{tan } \theta}{1 + \text{tan } \theta} \) and simplify it.

🎯 Exam Tip: When dealing with fractional expressions involving trigonometric ratios, first find all necessary individual ratios (like tangent in this case) and then substitute them into the main expression.

 

Question 2. In any triangle ABC, if \( \angle B = 90^\circ \) and side AB = 4 cm and AC = 5 cm, then find side BC.
Answer:
Given:
In \( \triangle ABC \), \( \angle B = 90^\circ \)
AB = 4 cm
AC = 5 cm
We use the Pythagorean theorem (also called Baudhayana formula) for a right-angled triangle:
\( \text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2 \)
\( \implies \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)
\( \implies \text{BC}^2 = \text{AC}^2 - \text{AB}^2 \)
\( \implies \text{BC} = \sqrt{\text{AC}^2 - \text{AB}^2} \)
\( \implies \text{BC} = \sqrt{5^2 - 4^2} \)
\( \implies \text{BC} = \sqrt{25 - 16} \)
\( \implies \text{BC} = \sqrt{9} \)
\( \implies \text{BC} = 3 \text{ cm} \)
In simple words: Since it's a right-angled triangle, we can use the Pythagorean theorem. We know the longest side (AC) and one other side (AB). To find the missing side (BC), we subtract the square of AB from the square of AC and then take the square root of the result.

🎯 Exam Tip: For right-angled triangle problems, always check if you can use the Pythagorean theorem \( (a^2 + b^2 = c^2) \) to find missing side lengths. Remember that 'c' is always the hypotenuse.

 

Question 3. Prove that \( \frac { 1 }{ 1 + \text{sin } \theta } + \frac { 1 }{ 1 - \text{sin } \theta } = 2 \text{ sec}^2 \theta \).
Answer:
L.H.S. \( = \frac { 1 }{ 1 + \text{sin } \theta } + \frac { 1 }{ 1 - \text{sin } \theta } \)
\( = \frac { (1 - \text{sin } \theta) + (1 + \text{sin } \theta) }{ (1 + \text{sin } \theta)(1 - \text{sin } \theta) } \) (Taking common denominator)
\( = \frac { 1 - \text{sin } \theta + 1 + \text{sin } \theta }{ 1^2 - \text{sin}^2 \theta } \) (Using \( (A+B)(A-B) = A^2-B^2 \))
\( = \frac { 2 }{ 1 - \text{sin}^2 \theta } \)
\( = \frac { 2 }{ \text{cos}^2 \theta } \) (Since \( 1 - \text{sin}^2 \theta = \text{cos}^2 \theta \))
\( = 2 \text{ sec}^2 \theta \) (Since \( \frac{1}{\text{cos } \theta} = \text{sec } \theta \))
\( = \text{R.H.S.} \) Hence proved. This identity shows how combining fractions can lead to simpler trigonometric expressions.
In simple words: To add the fractions, find a common bottom part. Multiply the first fraction's top and bottom by \( (1 - \text{sin } \theta) \) and the second by \( (1 + \text{sin } \theta) \). The bottom part becomes \( (1 - \text{sin}^2 \theta) \), which is \( \text{cos}^2 \theta \). The top part simplifies to 2. So we get \( \frac{2}{\text{cos}^2 \theta} \), which is \( 2 \text{ sec}^2 \theta \).

🎯 Exam Tip: When adding or subtracting trigonometric fractions, always find a common denominator and use algebraic identities (like difference of squares) to simplify the denominator first.

 

Question 4. Prove that \( \text{sec A (1 - sin A)(sec A + tan A) = 1} \).
Answer:
L.H.S. \( = \text{sec A (1 - sin A)(sec A + tan A)} \)
\( = \frac{1}{\text{cos A}} (1 - \text{sin A}) (\frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}) \) (Convert sec and tan to sin and cos)
\( = \frac{(1 - \text{sin A})}{\text{cos A}} \left( \frac{1 + \text{sin A}}{\text{cos A}} \right) \)
\( = \frac{(1 - \text{sin A})(1 + \text{sin A})}{\text{cos A} \cdot \text{cos A}} \)
\( = \frac{1^2 - \text{sin}^2 \text{A}}{\text{cos}^2 \text{A}} \) (Using \( (A-B)(A+B) = A^2-B^2 \))
\( = \frac{\text{cos}^2 \text{A}}{\text{cos}^2 \text{A}} \) (Since \( 1 - \text{sin}^2 \text{A} = \text{cos}^2 \text{A} \))
\( = 1 \)
\( = \text{R.H.S.} \) Hence proved. This identity simplifies by converting all terms to sine and cosine.
In simple words: First, change all \( \text{sec A} \) to \( \frac{1}{\text{cos A}} \) and all \( \text{tan A} \) to \( \frac{\text{sin A}}{\text{cos A}} \). Then, combine the terms. You will see that \( (1 - \text{sin A})(1 + \text{sin A}) \) becomes \( 1 - \text{sin}^2 \text{A} \), which is \( \text{cos}^2 \text{A} \). This will cancel out with the \( \text{cos}^2 \text{A} \) in the denominator, leaving 1.

🎯 Exam Tip: When proving identities involving products, converting everything to sine and cosine is often the most reliable strategy.

 

Question 5. Prove that \( (\text{sec } \theta - \text{tan } \theta)^2 = \frac { 1 - \text{sin } \theta }{ 1 + \text{sin } \theta } \).
Answer:
L.H.S. \( = (\text{sec } \theta - \text{tan } \theta)^2 \)
\( = \left( \frac{1}{\text{cos } \theta} - \frac{\text{sin } \theta}{\text{cos } \theta} \right)^2 \) (Convert sec and tan to sin and cos)
\( = \left( \frac{1 - \text{sin } \theta}{\text{cos } \theta} \right)^2 \)
\( = \frac{(1 - \text{sin } \theta)^2}{\text{cos}^2 \theta} \)
\( = \frac{(1 - \text{sin } \theta)^2}{1 - \text{sin}^2 \theta} \) (Since \( \text{cos}^2 \theta = 1 - \text{sin}^2 \theta \))
\( = \frac{(1 - \text{sin } \theta)(1 - \text{sin } \theta)}{(1 - \text{sin } \theta)(1 + \text{sin } \theta)} \) (Factorize the denominator using \( A^2 - B^2 = (A-B)(A+B) \))
\( = \frac { 1 - \text{sin } \theta }{ 1 + \text{sin } \theta } \)
\( = \text{R.H.S.} \) Hence proved. This proof demonstrates how to manipulate squared trigonometric expressions.
In simple words: Change \( \text{sec } \theta \) and \( \text{tan } \theta \) into their \( \text{sin } \theta \) and \( \text{cos } \theta \) forms. Combine the terms inside the bracket. Square the whole expression. Then, change \( \text{cos}^2 \theta \) at the bottom to \( 1 - \text{sin}^2 \theta \). Factor the bottom part and cancel out the common terms to get the right side.

🎯 Exam Tip: When proving identities involving squares of binomials, expand them first or convert to sine/cosine form to find common factors for cancellation.

 

Question 6. Prove that \( \frac { 1 }{ \text{cosec A - cot A} } - \frac { 1 }{ \text{sin A} } = \text{cot A} \).
Answer:
L.H.S. \( = \frac { 1 }{ \text{cosec A - cot A} } - \frac { 1 }{ \text{sin A} } \)
\( = \frac { \text{cosec}^2 \text{A} - \text{cot}^2 \text{A} }{ \text{cosec A - cot A} } - \frac { 1 }{ \text{sin A} } \) (Since \( \text{cosec}^2 \text{A} - \text{cot}^2 \text{A} = 1 \), replace 1 in the numerator)
\( = \frac { (\text{cosec A - cot A})(\text{cosec A + cot A}) }{ \text{cosec A - cot A} } - \frac { 1 }{ \text{sin A} } \) (Factorize numerator)
\( = \text{cosec A + cot A} - \frac { 1 }{ \text{sin A} } \)
\( = \text{cosec A + cot A} - \text{cosec A} \)
\( = \text{cot A} \)
\( = \text{R.H.S.} \) Hence proved. This identity uses the conjugate multiplication trick and reciprocal functions.
In simple words: For the first fraction, multiply the top and bottom by \( (\text{cosec A + cot A}) \). The bottom becomes \( \text{cosec}^2 \text{A} - \text{cot}^2 \text{A} \), which is 1. So, the first fraction simplifies to \( (\text{cosec A + cot A}) \). Then, \( \frac{1}{\text{sin A}} \) is \( \text{cosec A} \). Subtracting \( \text{cosec A} \) from \( (\text{cosec A + cot A}) \) leaves \( \text{cot A} \).

🎯 Exam Tip: When a denominator has a form like \( (A-B) \), consider multiplying the numerator and denominator by its conjugate \( (A+B) \) to simplify, especially if \( (A^2-B^2) \) is a known identity (like \( \text{cosec}^2 \text{A} - \text{cot}^2 \text{A} = 1 \)).

 

Question 7. Prove that \( \text{tan}^2 \phi + \text{cot}^2 \phi + 2 = \text{sec}^2 \phi \text{ cosec}^2 \phi \).
Answer:
L.H.S. \( = \text{tan}^2 \phi + \text{cot}^2 \phi + 2 \)
\( = ( \text{tan } \phi + \text{cot } \phi )^2 \) (Using \( (a+b)^2 = a^2+b^2+2ab \), where \( ab = \text{tan } \phi \cdot \text{cot } \phi = 1 \))
\( = \left( \frac{\text{sin } \phi}{\text{cos } \phi} + \frac{\text{cos } \phi}{\text{sin } \phi} \right)^2 \)
\( = \left( \frac{\text{sin}^2 \phi + \text{cos}^2 \phi}{\text{sin } \phi \text{ cos } \phi} \right)^2 \)
\( = \left( \frac{1}{\text{sin } \phi \text{ cos } \phi} \right)^2 \) (Since \( \text{sin}^2 \phi + \text{cos}^2 \phi = 1 \))
\( = \frac{1}{\text{sin}^2 \phi \text{ cos}^2 \phi} \)
\( = \text{cosec}^2 \phi \text{ sec}^2 \phi \)
\( = \text{R.H.S.} \) Hence proved. This identity shows a clever use of the \( (a+b)^2 \) expansion.
In simple words: The left side is similar to \( a^2 + b^2 + 2 \). Since \( \text{tan } \phi \cdot \text{cot } \phi = 1 \), this is actually \( (\text{tan } \phi + \text{cot } \phi)^2 \). Now, change \( \text{tan } \phi \) and \( \text{cot } \phi \) to sine and cosine. Add them, and the top part becomes 1. Then square the whole thing to get \( \text{sec}^2 \phi \text{ cosec}^2 \phi \).

🎯 Exam Tip: Look for patterns like \( a^2+b^2+2ab \) or \( a^2+b^2-2ab \) in trigonometric expressions, as they can often be simplified using \( (a+b)^2 \) or \( (a-b)^2 \).

 

Question 8. If \( \text{sin x} + \text{sin}^2 \text{x} = 1 \), show that \( \text{cos}^2 \text{x} + \text{cos}^4 \text{x} = 1 \).
Answer:
Given: \( \text{sin x} + \text{sin}^2 \text{x} = 1 \)
\( \implies \text{sin x} = 1 - \text{sin}^2 \text{x} \)
\( \implies \text{sin x} = \text{cos}^2 \text{x} \) (Using \( 1 - \text{sin}^2 \text{x} = \text{cos}^2 \text{x} \))
Now, square both sides of this equation:
\( \implies (\text{sin x})^2 = (\text{cos}^2 \text{x})^2 \)
\( \implies \text{sin}^2 \text{x} = \text{cos}^4 \text{x} \)
We know that \( \text{sin}^2 \text{x} = 1 - \text{cos}^2 \text{x} \). Substitute this into the equation above:
\( \implies 1 - \text{cos}^2 \text{x} = \text{cos}^4 \text{x} \)
\( \implies 1 = \text{cos}^2 \text{x} + \text{cos}^4 \text{x} \)
Hence proved. This problem elegantly uses the fundamental Pythagorean identity.
In simple words: From the first equation, we can say that \( \text{sin x} \) is equal to \( 1 - \text{sin}^2 \text{x} \), which is \( \text{cos}^2 \text{x} \). Now, if we square both sides, \( \text{sin}^2 \text{x} \) becomes \( \text{cos}^4 \text{x} \). Since \( \text{sin}^2 \text{x} \) is also \( 1 - \text{cos}^2 \text{x} \), we can replace it and move \( \text{cos}^2 \text{x} \) to the other side to get the desired result.

🎯 Exam Tip: When given a condition like \( \text{sin x} + \text{sin}^2 \text{x} = 1 \), try to express one trigonometric function in terms of another using identities, then substitute or square to manipulate the expression towards the desired proof.

 

Question 9. Prove that \( \text{cos}^2 \theta (1 + \text{tan}^2 \theta) + \text{sin}^2 \theta (1 + \text{cot}^2 \theta) = 2 \).
Answer:
L.H.S. \( = \text{cos}^2 \theta (1 + \text{tan}^2 \theta) + \text{sin}^2 \theta (1 + \text{cot}^2 \theta) \)
\( = \text{cos}^2 \theta (\text{sec}^2 \theta) + \text{sin}^2 \theta (\text{cosec}^2 \theta) \) (Using \( 1 + \text{tan}^2 \theta = \text{sec}^2 \theta \) and \( 1 + \text{cot}^2 \theta = \text{cosec}^2 \theta \))
\( = \text{cos}^2 \theta \cdot \frac{1}{\text{cos}^2 \theta} + \text{sin}^2 \theta \cdot \frac{1}{\text{sin}^2 \theta} \) (Using reciprocal identities for sec and cosec)
\( = 1 + 1 \)
\( = 2 \)
\( = \text{R.H.S.} \) Hence proved. This identity simplifies nicely by applying basic identities for tangent, cotangent, secant, and cosecant.
In simple words: First, change \( (1 + \text{tan}^2 \theta) \) to \( \text{sec}^2 \theta \) and \( (1 + \text{cot}^2 \theta) \) to \( \text{cosec}^2 \theta \). Then, \( \text{cos}^2 \theta \cdot \text{sec}^2 \theta \) becomes 1 because they are reciprocals. Similarly, \( \text{sin}^2 \theta \cdot \text{cosec}^2 \theta \) also becomes 1. Adding 1 and 1 gives 2.

🎯 Exam Tip: Always look for opportunities to replace \( (1 + \text{tan}^2 \theta) \) with \( \text{sec}^2 \theta \) and \( (1 + \text{cot}^2 \theta) \) with \( \text{cosec}^2 \theta \) as these are common simplification steps.

 

Question 10. If \( \text{sin } \theta + \text{cos } \theta = \text{a} \) and \( \text{sin } \theta - \text{cos } \theta = \text{b} \) then prove that \( \text{a}^2 + \text{b}^2 = 2 \).
Answer:
Given:
\( \text{sin } \theta + \text{cos } \theta = \text{a} \) ...(i)
\( \text{sin } \theta - \text{cos } \theta = \text{b} \) ...(ii)
Square both equations (i) and (ii):
From (i): \( (\text{sin } \theta + \text{cos } \theta)^2 = \text{a}^2 \)
\( \implies \text{sin}^2 \theta + \text{cos}^2 \theta + 2\text{sin } \theta \text{cos } \theta = \text{a}^2 \)
\( \implies 1 + 2\text{sin } \theta \text{cos } \theta = \text{a}^2 \) ...(iii) (Since \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \))
From (ii): \( (\text{sin } \theta - \text{cos } \theta)^2 = \text{b}^2 \)
\( \implies \text{sin}^2 \theta + \text{cos}^2 \theta - 2\text{sin } \theta \text{cos } \theta = \text{b}^2 \)
\( \implies 1 - 2\text{sin } \theta \text{cos } \theta = \text{b}^2 \) ...(iv)
Now, add equations (iii) and (iv):
\( (1 + 2\text{sin } \theta \text{cos } \theta) + (1 - 2\text{sin } \theta \text{cos } \theta) = \text{a}^2 + \text{b}^2 \)
\( \implies 1 + 2\text{sin } \theta \text{cos } \theta + 1 - 2\text{sin } \theta \text{cos } \theta = \text{a}^2 + \text{b}^2 \)
\( \implies 2 = \text{a}^2 + \text{b}^2 \)
Hence proved. This problem showcases how squaring and adding trigonometric expressions can simplify them.
In simple words: We are given two equations. Square both of them separately. When you square \( (\text{sin } \theta + \text{cos } \theta) \), you get \( 1 + 2\text{sin } \theta \text{cos } \theta \). When you square \( (\text{sin } \theta - \text{cos } \theta) \), you get \( 1 - 2\text{sin } \theta \text{cos } \theta \). Now, add these two new equations. The \( 2\text{sin } \theta \text{cos } \theta \) terms will cancel each other out, leaving \( 1 + 1 = 2 \), which equals \( \text{a}^2 + \text{b}^2 \).

🎯 Exam Tip: When given sums and differences of trigonometric functions, squaring both equations and then adding or subtracting them is a common technique to eliminate cross-terms and utilize identities.

Long Answer Type Questions

 

Question 1. Prove that \( (\text{sec A - cos A})(\text{cosec A - sin A}) = \frac { 1 }{ \text{tan A + cot A} } \).
Answer:
L.H.S. \( = (\text{sec A - cos A})(\text{cosec A - sin A}) \)
\( = \left( \frac{1}{\text{cos A}} - \text{cos A} \right) \left( \frac{1}{\text{sin A}} - \text{sin A} \right) \) (Convert sec and cosec to sin and cos)
\( = \left( \frac{1 - \text{cos}^2 \text{A}}{\text{cos A}} \right) \left( \frac{1 - \text{sin}^2 \text{A}}{\text{sin A}} \right) \)
\( = \left( \frac{\text{sin}^2 \text{A}}{\text{cos A}} \right) \left( \frac{\text{cos}^2 \text{A}}{\text{sin A}} \right) \) (Using \( 1 - \text{cos}^2 \text{A} = \text{sin}^2 \text{A} \) and \( 1 - \text{sin}^2 \text{A} = \text{cos}^2 \text{A} \))
\( = \text{sin A} \cdot \text{cos A} \)
Now, let's simplify the R.H.S.:
R.H.S. \( = \frac { 1 }{ \text{tan A + cot A} } \)
\( = \frac { 1 }{ \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} } \)
\( = \frac { 1 }{ \frac{\text{sin}^2 \text{A} + \text{cos}^2 \text{A}}{\text{sin A cos A}} } \)
\( = \frac { 1 }{ \frac{1}{\text{sin A cos A}} } \) (Since \( \text{sin}^2 \text{A} + \text{cos}^2 \text{A} = 1 \))
\( = \text{sin A cos A} \)
Since L.H.S. = R.H.S., the identity is proved. This problem is best solved by simplifying both sides independently.
In simple words: First, work with the left side. Change everything to sine and cosine. Simplify the terms inside the brackets. You will get \( \text{sin A cos A} \). Then, work with the right side. Change \( \text{tan A} \) and \( \text{cot A} \) to sine and cosine. Simplify the fraction at the bottom. You will again get \( \text{sin A cos A} \). Since both sides become the same, the proof is complete.

🎯 Exam Tip: When proving complex identities, it's often easier to simplify both the L.H.S. and R.H.S. independently until they match, rather than trying to transform one side directly into the other.

 

Question 2. Prove that \( \text{sin } \theta (1 + \text{tan } \theta) + \text{cos } \theta (1 + \text{cot } \theta) = \text{sec } \theta + \text{cosec } \theta \).
Answer:
L.H.S. \( = \text{sin } \theta (1 + \text{tan } \theta) + \text{cos } \theta (1 + \text{cot } \theta) \)
\( = \text{sin } \theta \left( 1 + \frac{\text{sin } \theta}{\text{cos } \theta} \right) + \text{cos } \theta \left( 1 + \frac{\text{cos } \theta}{\text{sin } \theta} \right) \) (Convert tan and cot to sin and cos)
\( = \text{sin } \theta \left( \frac{\text{cos } \theta + \text{sin } \theta}{\text{cos } \theta} \right) + \text{cos } \theta \left( \frac{\text{sin } \theta + \text{cos } \theta}{\text{sin } \theta} \right) \)
\( = (\text{sin } \theta + \text{cos } \theta) \left( \frac{\text{sin } \theta}{\text{cos } \theta} + \frac{\text{cos } \theta}{\text{sin } \theta} \right) \) (Factor out \( (\text{sin } \theta + \text{cos } \theta) \))
\( = (\text{sin } \theta + \text{cos } \theta) \left( \frac{\text{sin}^2 \theta + \text{cos}^2 \theta}{\text{sin } \theta \text{cos } \theta} \right) \)
\( = (\text{sin } \theta + \text{cos } \theta) \left( \frac{1}{\text{sin } \theta \text{cos } \theta} \right) \) (Since \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \))
\( = \frac{\text{sin } \theta}{\text{sin } \theta \text{cos } \theta} + \frac{\text{cos } \theta}{\text{sin } \theta \text{cos } \theta} \)
\( = \frac{1}{\text{cos } \theta} + \frac{1}{\text{sin } \theta} \)
\( = \text{sec } \theta + \text{cosec } \theta \)
\( = \text{R.H.S.} \) Hence proved. This identity shows the power of factoring common terms.
In simple words: First, change \( \text{tan } \theta \) to \( \frac{\text{sin } \theta}{\text{cos } \theta} \) and \( \text{cot } \theta \) to \( \frac{\text{cos } \theta}{\text{sin } \theta} \). Then, simplify inside the brackets by finding common denominators. You will find that \( (\text{sin } \theta + \text{cos } \theta) \) is a common factor. Take it out and simplify the remaining terms. This leads to \( \frac{1}{\text{cos } \theta} + \frac{1}{\text{sin } \theta} \), which is \( \text{sec } \theta + \text{cosec } \theta \).

🎯 Exam Tip: When terms like \( (1 + \text{tan } \theta) \) and \( (1 + \text{cot } \theta) \) appear, converting to sine and cosine and looking for common factors is a very effective strategy for simplification.

 

Question 3. Prove that \( \frac { 1 }{ \text{sec A + tan A} } - \frac { 1 }{ \text{cos A} } = \frac { 1 }{ \text{cos A} } - \frac { 1 }{ \text{sec A - tan A} } \).
Answer:
Let's rearrange the terms to bring similar expressions together:
\( \frac { 1 }{ \text{sec A + tan A} } + \frac { 1 }{ \text{sec A - tan A} } = \frac { 1 }{ \text{cos A} } + \frac { 1 }{ \text{cos A} } \)
L.H.S. \( = \frac { 1 }{ \text{sec A + tan A} } + \frac { 1 }{ \text{sec A - tan A} } \)
\( = \frac { (\text{sec A - tan A}) + (\text{sec A + tan A}) }{ (\text{sec A + tan A})(\text{sec A - tan A}) } \)
\( = \frac { \text{sec A - tan A + sec A + tan A} }{ \text{sec}^2 \text{A} - \text{tan}^2 \text{A} } \) (Using \( (A+B)(A-B) = A^2-B^2 \))
\( = \frac { 2\text{sec A} }{ 1 } \) (Since \( \text{sec}^2 \text{A} - \text{tan}^2 \text{A} = 1 \))
\( = 2\text{sec A} \)
R.H.S. \( = \frac { 1 }{ \text{cos A} } + \frac { 1 }{ \text{cos A} } \)
\( = \frac { 2 }{ \text{cos A} } \)
\( = 2\text{sec A} \) (Since \( \frac{1}{\text{cos A}} = \text{sec A} \))
Since L.H.S. = R.H.S., the identity is proved. This problem is simpler by rearranging terms first.
In simple words: First, move the terms around so that the fractions with \( \text{sec A} \) and \( \text{tan A} \) are on one side, and the \( \text{cos A} \) terms are on the other. Add the fractions on the left side by finding a common denominator. The bottom part will become \( \text{sec}^2 \text{A} - \text{tan}^2 \text{A} \), which is 1. The top part simplifies to \( 2\text{sec A} \). On the right side, \( \frac{1}{\text{cos A}} + \frac{1}{\text{cos A}} \) is \( \frac{2}{\text{cos A}} \), which is also \( 2\text{sec A} \). Both sides match.

🎯 Exam Tip: Sometimes, rearranging the equation before starting the proof can make it significantly simpler. Look for terms that can be combined or simplified with an identity on each side.

 

Question 4. Prove that \( \frac { 1 + \text{cos } \theta - \text{sin}^2 \theta }{ \text{sin } \theta (1 + \text{cos } \theta) } = \text{cot } \theta \).
Answer:
L.H.S. \( = \frac { 1 + \text{cos } \theta - \text{sin}^2 \theta }{ \text{sin } \theta (1 + \text{cos } \theta) } \)
\( = \frac { (1 - \text{sin}^2 \theta) + \text{cos } \theta }{ \text{sin } \theta (1 + \text{cos } \theta) } \) (Rearrange terms in numerator)
\( = \frac { \text{cos}^2 \theta + \text{cos } \theta }{ \text{sin } \theta (1 + \text{cos } \theta) } \) (Since \( 1 - \text{sin}^2 \theta = \text{cos}^2 \theta \))
\( = \frac { \text{cos } \theta ( \text{cos } \theta + 1 ) }{ \text{sin } \theta (1 + \text{cos } \theta) } \) (Factor out \( \text{cos } \theta \) from numerator)
\( = \frac { \text{cos } \theta }{ \text{sin } \theta } \) (Cancel out \( (1 + \text{cos } \theta) \) from numerator and denominator)
\( = \text{cot } \theta \)
\( = \text{R.H.S.} \) Hence proved. This identity simplifies by substituting a key identity and factoring.
In simple words: Look at the top part of the fraction. Change \( (1 - \text{sin}^2 \theta) \) to \( \text{cos}^2 \theta \). Then, you will have \( \text{cos}^2 \theta + \text{cos } \theta \) at the top. Take out \( \text{cos } \theta \) as a common factor, leaving \( \text{cos } \theta ( \text{cos } \theta + 1 ) \). The \( ( \text{cos } \theta + 1 ) \) will cancel with the \( (1 + \text{cos } \theta) \) at the bottom. This leaves \( \frac{\text{cos } \theta}{\text{sin } \theta} \), which is \( \text{cot } \theta \).

🎯 Exam Tip: Always look for ways to factorize the numerator and denominator after applying identities. Common factors often lead to significant simplification.

 

Question 5. Prove that \( \text{sec}^6 \theta = \text{tan}^6 \theta + 3 \text{ tan}^2 \theta \text{ sec}^2 \theta + 1 \).
Answer:
We know the identity \( \text{sec}^2 \theta = 1 + \text{tan}^2 \theta \).
Cube both sides:
\( (\text{sec}^2 \theta)^3 = (1 + \text{tan}^2 \theta)^3 \)
\( \implies \text{sec}^6 \theta = 1^3 + (\text{tan}^2 \theta)^3 + 3 \cdot 1 \cdot \text{tan}^2 \theta (1 + \text{tan}^2 \theta) \) (Using \( (A+B)^3 = A^3 + B^3 + 3AB(A+B) \))
\( \implies \text{sec}^6 \theta = 1 + \text{tan}^6 \theta + 3 \text{ tan}^2 \theta (\text{sec}^2 \theta) \) (Substitute \( (1 + \text{tan}^2 \theta) \) with \( \text{sec}^2 \theta \))
\( \implies \text{sec}^6 \theta = \text{tan}^6 \theta + 3 \text{ tan}^2 \theta \text{ sec}^2 \theta + 1 \)
Hence proved. This identity is derived by cubing a basic trigonometric identity.
In simple words: Start with the basic rule \( \text{sec}^2 \theta = 1 + \text{tan}^2 \theta \). Now, cube both sides of this equation. Use the algebraic formula for \( (a+b)^3 \). When you expand it, you will see \( (1 + \text{tan}^2 \theta) \) appears, which you can change back to \( \text{sec}^2 \theta \). This will give you the required proof.

🎯 Exam Tip: When proving identities involving higher powers, consider if cubing or squaring a known identity (like \( \text{sec}^2 \theta = 1 + \text{tan}^2 \theta \)) can lead to the desired expression.

 

Question 6. Prove that \( \frac { \text{tan } \theta }{ \text{sec } \theta - 1 } + \frac { \text{tan } \theta }{ \text{sec } \theta + 1 } = 2 \text{ cosec } \theta \).
Answer:
L.H.S. \( = \frac { \text{tan } \theta }{ \text{sec } \theta - 1 } + \frac { \text{tan } \theta }{ \text{sec } \theta + 1 } \)
\( = \text{tan } \theta \left( \frac { 1 }{ \text{sec } \theta - 1 } + \frac { 1 }{ \text{sec } \theta + 1 } \right) \) (Factor out \( \text{tan } \theta \))
\( = \text{tan } \theta \left( \frac { (\text{sec } \theta + 1) + (\text{sec } \theta - 1) }{ (\text{sec } \theta - 1)(\text{sec } \theta + 1) } \right) \)
\( = \text{tan } \theta \left( \frac { \text{sec } \theta + 1 + \text{sec } \theta - 1 }{ \text{sec}^2 \theta - 1^2 } \right) \) (Using \( (A-B)(A+B) = A^2-B^2 \))
\( = \text{tan } \theta \left( \frac { 2\text{sec } \theta }{ \text{sec}^2 \theta - 1 } \right) \)
\( = \text{tan } \theta \left( \frac { 2\text{sec } \theta }{ \text{tan}^2 \theta } \right) \) (Since \( \text{sec}^2 \theta - 1 = \text{tan}^2 \theta \))
\( = \frac { \text{tan } \theta \cdot 2\text{sec } \theta }{ \text{tan}^2 \theta } \)
\( = \frac { 2\text{sec } \theta }{ \text{tan } \theta } \)
\( = \frac { 2 \cdot \frac{1}{\text{cos } \theta} }{ \frac{\text{sin } \theta}{\text{cos } \theta} } \) (Convert sec and tan to sin and cos)
\( = \frac { 2 }{ \text{cos } \theta } \cdot \frac { \text{cos } \theta }{ \text{sin } \theta } \)
\( = \frac { 2 }{ \text{sin } \theta } \)
\( = 2 \text{ cosec } \theta \)
\( = \text{R.H.S.} \) Hence proved. This identity simplifies by factoring, finding a common denominator, and using reciprocal relations.
In simple words: First, take out \( \text{tan } \theta \) as a common factor. Then, add the two fractions inside the bracket. The bottom part becomes \( \text{sec}^2 \theta - 1 \), which is \( \text{tan}^2 \theta \). The top part is \( 2\text{sec } \theta \). So, you get \( \text{tan } \theta \cdot \frac{2\text{sec } \theta}{\text{tan}^2 \theta} \). One \( \text{tan } \theta \) cancels out, leaving \( \frac{2\text{sec } \theta}{\text{tan } \theta} \). Convert \( \text{sec } \theta \) and \( \text{tan } \theta \) to sine and cosine. Simplify to get \( 2 \text{ cosec } \theta \).

🎯 Exam Tip: When fractions share a common factor in the numerator, factor it out first. Then, combine the remaining fractions by finding a common denominator. This often simplifies the problem significantly.

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