ssc gd trigonometry

• Introduction to Trigonometry
• Understanding Trigonometric Ratios (SOH CAH TOA)
  • The Six Ratios
  • Solved Examples for Ratios
• Standard Values (0°, 30°, 45°, 60°, 90°)
  • The Important Value Table
  • Solved Examples using Standard Values
• Simple Trigonometric Identities
  • The Three Main Rules
  • Solved Examples using Identities

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Introduction to Trigonometry

Trigonometry is a big word, but it just means measuring triangles.

Imagine you are standing far away from a tall building. You want to know how high the building is, but you cannot climb it with a measuring tape.

• Trigonometry helps us find the height using angles and distances on the ground.
• It only works with right-angled triangles (triangles that have one ${90}^{\circ }$ corner).
• Learning these basic rules is key to solving ssc gd trigonometry questions quickly.

Understanding Trigonometric Ratios (SOH CAH TOA)

In a right-angled triangle, we give special names to the three sides based on the angle ($\theta$) we are looking at.

• Hypotenuse (H): This is the longest side. It is always opposite the ${90}^{\circ }$ angle.
• Opposite (O): This side is opposite the angle ($\theta$) we are using.
• Adjacent (A): This side is next to the angle ($\theta$) we are using (but it is not the hypotenuse).

The Six Ratios

We use three main ratios (fractions) to connect the sides and the angle. We remember them using the simple phrase: SOH CAH TOA.

Key Relationships to Remember:

• $\csc \theta =1/\sin \theta$
• $\sec \theta =1/\cos \theta$
• $\cot \theta =1/\tan \theta$
• $\tan \theta =\sin \theta /\cos \theta$

Solved Examples for Ratios

These types of ssc gd trigonometry questions test your understanding of the basic side ratios.

Example 1: Finding Sine and Cosine

In a right-angled triangle, the Opposite side is 3 cm and the Hypotenuse is 5 cm. Find the value of $\sin \theta$.

Solution:

1. First, let’s understand what we need to find. We need $\sin \theta$.
2. Now, remember the SOH rule: $\sin \theta =\text{Opposite}/\text{Hypotenuse}$.
3. Put the numbers in. Opposite $=3$, Hypotenuse $=5$. $$\sin \theta =3/5$$
4. Calculate it. $$\sin \theta =0.6$$
5. So, the answer is $3/5$.

Example 2: Finding Tangent

If the Adjacent side is 12 and the Opposite side is 5, what is $\tan \theta$?

Solution:

1. We need $\tan \theta$. We use the TOA rule.
2. The formula is: $\tan \theta =\text{Opposite}/\text{Adjacent}$.
3. Put the numbers in. Opposite $=5$, Adjacent $=12$. $$\tan \theta =5/12$$
4. Since 5 and 12 cannot be simplified, this is our final answer.
5. So, the answer is $5/12$.

Example 3: Using Reciprocals

If $\cos \theta =4/5$, find the value of $\sec \theta$.

Solution:

1. First, we remember the relationship between $\cos \theta$ and $\sec \theta$. They are reciprocals (flips).
2. The formula is: $\sec \theta =1/\cos \theta$.
3. Now, write down the value of $\cos \theta$. $\cos \theta =4/5$.
4. To find $\sec \theta$, we just flip the fraction $4/5$. $$\sec \theta =5/4$$
5. So, the answer is $5/4$. These are common ssc gd trigonometry questions.

Example 4: Finding Sine when Cosine is Known

If $\cos A=3/5$, find $\tan A$. (Hint: Use Pythagoras Theorem to find the missing side).

Solution:

1. We know $\cos A=\text{Adjacent}/\text{Hypotenuse}=3/5$. So, Adjacent $=3$ and Hypotenuse $=5$.
2. We need $\tan A=\text{Opposite}/\text{Adjacent}$. We must find the Opposite side (O).
3. Use Pythagoras Theorem: $(\text{Opposite}{)}^2+(\text{Adjacent}{)}^2=(\text{Hypotenuse}{)}^2$. $$O^2+3^2=5^2$$
4. Calculate the squares: $$O^2+9=25$$ $$O^2=25-9$$ $$O^2=16$$ $$O=4$$ (The Opposite side is 4).
5. Now find $\tan A$: $\tan A=\text{Opposite}/\text{Adjacent}=4/3$.
6. So, the answer is $4/3$.

Standard Values (0°, 30°, 45°, 60°, 90°)

In SSC GD exams, you must know the values of the ratios for certain special angles by heart. These are the most common ssc gd trigonometry questions asked.

The Important Value Table

You must memorize this table. It is the foundation for solving many problems.

• Tip: If you know the $\sin$ row, you know the $\cos$ row (it is just the $\sin$ row written backward).
• Tip: $\tan \theta =\sin \theta /\cos \theta$.

Solved Examples using Standard Values

These examples show how to use the table to solve ssc gd trigonometry questions.

Example 1: Simple Addition

Find the value of $\sin {30}^{\circ }+\cos {60}^{\circ }$.

Solution:

1. First, we look up the values in our table. $$\sin {30}^{\circ }=1/2$$ $$\cos {60}^{\circ }=1/2$$
2. Now, we put these numbers into the equation. $$\text{Value}=1/2+1/2$$
3. We add the fractions. Half plus half equals one whole. $$\text{Value}=2/2=1$$
4. So, the answer is 1.

Example 2: Multiplication

Calculate $2\tan {45}^{\circ }+\sin {90}^{\circ }$.

Solution:

1. First, find the values from the table. $$\tan {45}^{\circ }=1$$ $$\sin {90}^{\circ }=1$$
2. Now, substitute the values into the expression. $$\text{Value}=2\times (1)+1$$
3. Do the multiplication first (BODMAS rule). $$\text{Value}=2+1$$
4. Calculate the final sum. $$\text{Value}=3$$
5. So, the answer is 3.

Example 3: Using Square Roots

Find the value of ${\sin }^2{60}^{\circ }+{\cos }^2{30}^{\circ }$.

Solution:

1. First, find the values for $\sin {60}^{\circ }$ and $\cos {30}^{\circ }$. $$\sin {60}^{\circ }=\sqrt{3}/2$$ $$\cos {30}^{\circ }=\sqrt{3}/2$$
2. Now, we need to square these values. Remember, $(\sqrt{3}/2{)}^2$ means $(\sqrt{3}/2)\times (\sqrt{3}/2)$. $$(\sqrt{3}/2{)}^2=3/4$$
3. Substitute the squared values back into the equation. $$\text{Value}=3/4+3/4$$
4. Add the fractions. $$\text{Value}=6/4$$
5. Simplify the fraction by dividing the top and bottom by 2. $$\text{Value}=3/2$$

Example 4: Using Reciprocals and Division

Calculate $\frac{\tan {30}^{\circ }}{\cot {60}^{\circ }}$.

Solution:

1. First, find the values from the table. $$\tan {30}^{\circ }=1/\sqrt{3}$$ $$\cot {60}^{\circ }=1/\sqrt{3}$$ (Remember, $\cot$ is the flip of $\tan$).
2. Now, substitute the values into the fraction. $$\text{Value}=\frac{1/\sqrt{3}}{1/\sqrt{3}}$$
3. When the top number (numerator) and the bottom number (denominator) are exactly the same, the answer is 1. $$\text{Value}=1$$
4. So, the answer is 1. These are typical ssc gd trigonometry questions that look hard but are easy if you know the table.

Simple Trigonometric Identities

Identities are like special math rules that are always true, no matter what angle ($\theta$) you use. You must memorize these three main rules to solve advanced ssc gd trigonometry questions.

The Three Main Rules

These rules involve squaring the ratios.

1. Rule 1 (The most important): $${\sin }^2\theta +{\cos }^2\theta =1$$

   • Simple Meaning: If you square the sine of an angle and add it to the square of the cosine of the same angle, the answer is always 1.

2. Rule 2: $$1+{\tan }^2\theta ={\sec }^2\theta$$

   • Simple Meaning: This rule connects tangent and secant.

3. Rule 3: $$1+{\cot }^2\theta ={\csc }^2\theta$$

   • Simple Meaning: This rule connects cotangent and cosecant.

Solved Examples using Identities

Example 1: Using Rule 1 Directly

Simplify the expression: $5{\sin }^2{15}^{\circ }+5{\cos }^2{15}^{\circ }$.

Solution:

1. First, notice that the angle (${15}^{\circ }$) is the same for both ${\sin }^2$ and ${\cos }^2$.
2. We can take the number 5 out (factor it). $$\text{Expression}=5({\sin }^2{15}^{\circ }+{\cos }^2{15}^{\circ })$$
3. Now, we use Identity Rule 1: ${\sin }^2\theta +{\cos }^2\theta =1$. $$\text{Expression}=5\times (1)$$
4. Calculate the final value. $$\text{Expression}=5$$
5. So, the answer is 5.

Example 2: Using Rule 2

If $\tan \theta =3$, find the value of ${\sec }^2\theta$.

Solution:

1. First, we look for the rule that connects $\tan \theta$ and $\sec \theta$. That is Rule 2. $${\sec }^2\theta =1+{\tan }^2\theta$$
2. We are given $\tan \theta =3$. We need to find ${\tan }^2\theta$. $${\tan }^2\theta =3^2=9$$
3. Now, put the number 9 into the formula. $${\sec }^2\theta =1+9$$
4. Calculate the final sum. $${\sec }^2\theta =10$$
5. So, the answer is 10.

Example 3: Rearranging Rule 1

Simplify: ${\sin }^2{40}^{\circ }-1$.

Solution:

1. This looks tricky, but we start with Rule 1: ${\sin }^2\theta +{\cos }^2\theta =1$.
2. We want to find what ${\sin }^2\theta -1$ equals. Let’s move the numbers around in Rule 1. $${\sin }^2\theta -1=-{\cos }^2\theta$$
3. Now, we apply this to the question ($\theta ={40}^{\circ }$). $${\sin }^2{40}^{\circ }-1=-{\cos }^2{40}^{\circ }$$
4. So, the simplified expression is $-{\cos }^2{40}^{\circ }$.

Example 4: Using Rule 3

Simplify the expression: ${\csc }^2\theta -{\cot }^2\theta$. These are tricky ssc gd trigonometry questions.

Solution:

1. First, we look at Rule 3, which connects $\csc$ and $\cot$. $$1+{\cot }^2\theta ={\csc }^2\theta$$
2. We want to find ${\csc }^2\theta -{\cot }^2\theta$. We need to move ${\cot }^2\theta$ to the other side of the equals sign. $$1={\csc }^2\theta -{\cot }^2\theta$$
3. Since the expression we needed to simplify is exactly the same as the right side of our rearranged formula, the answer is 1. $$\text{Value}=1$$
4. So, the answer is 1.