Subject: Optional Maths

The ratios which are generally used to calculate the unknown lengths and angles in a right triangle is called trigonometric ratios. Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions. This note has information about trigonometric ratios, its operations, trigonometric expressions and reciprocals relations.

There are six trigonometric ratios which relate the sides of a right triangle to its angles. Trigonometric ratios are generally used to calculate the unknown lengths and angles in a right triangle. The six trigonometric ratios are tabulated below: -

1) Sinθ | \(\frac{perpendicuar}{hypotenuse}\) (\(\frac{p}{h}\)) |

2) Cosθ | \(\frac{base}{hypotenuse}\) (\(\frac{b}{h}\)) |

3) Tanθ | \(\frac{perpendicuar}{base}\) (\(\frac{p}{b}\)) |

4) Cosecθ | \(\frac{hypotenuse}{perpendicuar}\) (\(\frac{h}{p}\)) |

5) Secθ | \(\frac{hypotenuse}{base}\) (\(\frac{h}{b}\)) |

6) Cotθ | \(\frac{base}{perpendicuar}\) (\(\frac{b}{p}\)) |

**NOTE: - Since, Sinθ ≠ Sin×θ, Cosθ ≠ Cos×θ, etc.**

As we know that the trigonometric ratios are numbers like fraction, decimal or whole numbers. We can operate the operations of addition, subtraction, multiplication and division on them that we do in algebra.

Operations | In algebra | In Trigonometry |

Addition | 2a + 4a = 6a 4x ^{2} + 7x^{2} = 11x^{2} |
2sinθ + 4sinθ = 6sinθ 4cos ^{2}θ + 7cos^{2}θ = 11cos^{2}θ |

Subtraction | 11z − 6z = 5z 8x ^{2}− 5x^{2} = 3x^{2} |
11tanθ − 6tanθ = 5tanθ 8sec ^{2}θ − 5sec^{2}θ = 3sec^{2}θ |

Multiplication | 7a × 4a = 28a^{2}5y^{2}× 3y^{2}= 15y^{4} |
7sinθ × 4sinθ = 28sin^{2}θ5tan ^{2}θ × 4tan^{2}θ = 20tan^{4}θ |

Division | \(\frac{q^{5}}{q}\) = q^{4}10b ÷ 5b = 2 |
\(\frac{cosθ^{5}}{cosθ}\) = cosθ^{4}10secθ ÷ 5secθ = 2 |

Trigonometric Expressions

Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions.

We can simplify or factorise the trigonometric expressions as in algebra. For examples,

- (a+b) (a−b) = a
^{2}− b^{2}

(SinA + CosA) (SinA − CosA) = Sin^{2}A − Cos^{2}A - (a+b)
^{2}= a^{2}+ 2ab + b^{2}(SinA + CosA)^{2}= Sin^{2}A + 2SinA.CosA + Cos^{2}A - a
^{3}+b^{3}= (a+b) (a^{2}−ab+b^{2})

Sin^{3}A + Cos^{3}A = (SinA + CosA)(Sin^{2}A − 2SinA.CosA + Cos^{2}A) - a
^{3}−b^{3}= (a+b) (a^{2}+ab+b^{2})

Sin^{3}A − Cos^{3}A = (SinA − CosA) (Sin^{2}A + SinA.CosA + Cos^{2}A)

Relation between the Trigonometric Ratios of an Angle

Simply there are four fundamental relations of trigonometric ratios. They are,

- Reciprocal Relation
- Quotient Relation
- Pythagoras Relation
- Derived Relation

**Reciprocal Relation**

Reciprocal relations of trigonometric ratios are explained to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals. The reciprocals relations are given below: -

Sinθ = \(\frac{p}{h}\) and Cosecθ = \(\frac{h}{p}\)

Then, sinθ × cosecθ = \(\frac{p}{h}\) × \(\frac{h}{p}\) = 1

∴ Sinθ × cosecθ = 1 |

sinθ = \(\frac{1}{cosecθ }\), cosecθ = \(\frac{1}{sinθ }\)

Also, cosθ = \(\frac{b}{h}\) and secθ = \(\frac{h}{b}\)

Then, cosθ × secθ = \(\frac{b}{h}\) × \(\frac{h}{b}\) = 1

∴ cosθ × secθ = 1 |

cosθ = \(\frac{1}{secθ}\), secθ \(\frac{1}{cosθ}\)

And, tanθ = \(\frac{p}{b}\), cotθ = \(\frac{b}{p}\)

Then tanθ × cotθ = 1

∴ tanθ × cotθ = 1 |

tanθ = \(\frac{1}{cotθ}\), cotθ = \(\frac{1}{tanθ}\)

**Quotient Relation**

We have, sinθ = \(\frac{p}{h}\), cosθ = \(\frac{b}{h}\)

Then, \(\frac{sinθ}{cosθ}\) = \(\frac{p}{h}\) × \(\frac{h}{b}\) = \(\frac{p}{b}\)

We have tanθ = \(\frac{p}{b}\)**∴ tanθ = \(\frac{sinθ}{cosθ}\)**Similarly,

Pythagoras Relation

CA

or, h

Dividing both sides

or, \(\frac{h^{2}}{h^{2}}\) = \(\frac{p^{2}+b^{2}}{h^{2}}\)

or, 1 = \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\)

or, \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\) = 1

or, (\(\frac{p^{2}}{h^{2}}\)) + (\(\frac{b^{2}}{h^{2}}\)) = 1

or, (sinθ)

∴ sin

cos

Also,

sinθ = \(\sqrt{1-cos^{2}}{θ}\)

or, cosθ = \(\sqrt{1−sin^{2}}{θ}\)

Again,

p

or,h

Dividing both sides by b

or, \(\frac{h^{2}}{b{2}}\) - \(\frac{p^{2}}{b^{2}}\) = \(\frac{b^{2}}{b^{2}}\)

or, (\(\frac{h}{b}\))

or, (secθ)

or, sec

or, sec

or, tan

Also,

secθ = \(\sqrt{1+tan^{2}}{θ}\)

or, tanθ = \(\sqrt{sec^{2} θ − 1}\)

Again,

p

or, h

Dividing both sides by p

or, (cosecθ)

or, cosec

or, cosec

or, cot

Also, cotθ = \(\sqrt{cosec^{2} θ − 1}\)

or, cosecθ = \(\sqrt{1+cot^{2}}{θ}\)

To prove trigonometric identities, we may follow any one of the following: -

- Begin from the left-hand side (L.H.S.) and deduct it to the right-hand side (R.H.S.), if L.H.S. is more complex.
- Begin from R.H.S. and deduct it to L.H.S., if L.H.S. is complex.
- Reduce both the L.H.S. and the R.H.S. to the same expression if both the expression are complex.
- By transposition or cross multiplication, change the identity into the appropriate form. Then show that the new L.H.S. = the new R.H.S.

With the use of two methods, we can express the trigonometric ratios in terms of other ratios of the same angle. The two methods are as followings: -

- Using trigonometric relations i.e Trigonometric formulae.
- Using Pythagoras theorem.

NOTE: - We can also find remaining ratios if the value of any trigonometric ratio of an angle is given. |

Trigonometric Ratios of some angles

As trigonometry have different angles, for those different angles trigonometric ratios also have different values. In the trigonometry the angles 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° are taken as standard angles.

We can find the trigonometric ratios of 0° and 90°. The following table shows the values of trigonometric ratios of the angles 0°, 30°, 45°, 60° and 90°.

Angles^{→}↓ |
0° | 30° | 45° | 60° | 90° |

sinθ | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |

cosθ | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{1}{2}\) | 0 |

tanθ | 0 | \(\frac{1}{\sqrt3}\) | 1 | \(\sqrt{3}\) | ∞ |

cosecθ | ∞ | 2 | \(\sqrt{2}\) | \(\frac{2}{\sqrt3}\) | 1 |

secθ | 1 | \(\frac{2}{\sqrt3}\) | \(\sqrt{2}\) | 2 | ∞ |

cotθ | ∞ | \(\sqrt{3}\) | 1 | \(\frac{1}{\sqrt3}\) | 0 |

The standard angles i.e 0°, 30°, 45°, 60°, 90° in prompt way is given below: -

**Process 1: -** Put the number from 0 to 4 as follows: -

0° | 30° | 45° | 60° | 90° |

0 | 1 | 2 | 3 | 4 |

**Process 2: -**

Divide each number by 4

0° | 30° | 45° | 60° | 90° |

\(\frac{0}{4}\) | \(\frac{1}{4}\) | \(\frac{2}{4}\) | \(\frac{3}{4}\) | \(\frac{4}{4}\) |

**Process ****3: -** Take square root of all the numbers

0° | 30° | 45° | 60° | 90° |

\(\sqrt{\frac{0}{4}}\) | \(\sqrt{\frac{1}{4}}\) | \(\sqrt{\frac{2}{4}}\) | \(\sqrt{\frac{3}{4}}\) | \(\sqrt{\frac{4}{4}}\) |

**Process 4: -** The values which are obtained at first are the values of sine of the standard angles.

Angles | 0° | 30° | 45° | 60° | 90° |

sin | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |

**Process 5: -**Then reversing the order, the value of cosine of theb standard angles are obtained.

Angles | 0° | 30° | 45° | 60° | 90° |

cos | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt2}\) | \(\frac{1}{2}\) | 0 |

**Process: -**After dividing the value of sine by the value of cosine, the values of tangent can be obtained: -

Angle | 0° | 30° | 45° | 60° | 90° |

tan | 0 | \(\frac{1}{\sqrt3}\) | 1 | \(\sqrt{3}\) | ∞ |

NOTE: - The reciprocal of sine, cosine and tangent are the values of cosecant, secant and cotangent respectively. |

Complementary Angles

Two angles are called complementary when those angles are added up to 90° or the sum of two angles is 90°.

For example,

60° and 30° are complementary angles.

5° and 85° are complementary angles.

30° and 60°, 50° and 40°, (90° - θ) and θ are complementary angles.

Here ABC is a right angle triangle right angled at B.

Let, ∠ABC = θ, then ∠CAB = 90° - θ

Now taking θ as an angle of reference,

AB = Perpendicular (p)

BC = Base (b)

CA = Hyptenuse (h)

Sinθ = \(\frac{p}{h}\) = \(\frac{AB}{CA}\) .............. .(i)

Cosθ = \(\frac{b}{h}\) = \(\frac{BC}{CA}\) ..............(ii)

Tanθ = \(\frac{p}{b}\) = \(\frac{AB}{BC}\) ..............(iii)

Cotθ = \(\frac{b}{p}\) = \(\frac{BC}{AB}\) ...............(iv)

Taking (90° - θ) as an angle of reference,

BC = Perpendicular (p)

AB = Base (b)

CA = Hypotenuse (h)

Now,

Sinθ(90° - θ) = \(\frac{p}{h}\) = \(\frac{BC}{CA}\) = Cosθ

Cosθ(90° - θ) = \(\frac{b}{h}\) = \(\frac{AB}{CA}\) = Sinθ

Tanθ(90° - θ) = \(\frac{p}{b}\) = \(\frac{BC}{AB}\) = Cotθ

On the other hand,

cosec(90° - θ) = secθ

sec(90° - θ) = cosecθ

cot(90° - θ) = tanθ

(sinθ)^{2} + (cosθ)^{2} = 1 |
sec^{2}θ = 1+tan^{2}θ |
cosec^{2}θ - cot^{2}θ = 1 |

sin^{2}θ = 1-cos^{2}θ |
sec^{2}θ - tan^{2}θ =1 |
cosec^{2}θ = 1 + cot^{2}θ |

, sinθ= \(\sqrt{1-cos^{2}}{θ}\) | tan^{2}θ= sec^{2}θ - 1 |
cot^{2}θ = cosec^{2}θ - 1 |

cosθ = \(\sqrt{1-sin^{2}}{θ}\) |
secθ = \(\sqrt{1+tan^{2}}{θ}\) |
cotθ =\(\sqrt{cosec^{2} θ-1}\) |

sec^{2}θ - tan^{2}θ =1 |
tanθ = \(\sqrt{sec^{2} θ-1}\) | cosecθ = \(\sqrt{1+cot^{2}}{θ}\) |

- It includes every relationship which established among the people.
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- common interests and common objectives are not necessary for society.

Find the value of sinα, cosα and tanα from the given figure.

Solution:

From the given figure,

Hypotenuse (h) = 13cm

Perpendicular (p) = 5cm

Base (b) = 12cm

Now,

\(\therefore\) sinα = \(\frac{p}{h}\) = \(\frac{5}{13}\)

\(\therefore\) cosα = \(\frac{b}{h}\) = \(\frac{12}{13}\)

\(\therefore\) tanα = \(\frac{p}{b}\) = \(\frac{5}{12}\)

Find the value of cosecθ, secθ and cotθ from the given figure.

Solution:

From the given figure,

Hypotenuse (h) = 10cm

Perpendicular (p) = 6cm

Base (b) = 8cm

Now,

\(\therefore\) cosecθ = \(\frac{h}{p}\) = \(\frac{10}{6}\) = \(\frac{5}{3}\)

\(\therefore\) secθ = \(\frac{h}{b}\) = \(\frac{10}{8}\) = \(\frac{5}{4}\)

\(\therefore\) cotθ = \(\frac{b}{p}\) = \(\frac{8}{6}\) = \(\frac{4}{3}\)

Add: 3sinA + 7sinA + 10sinA

Solution:

= (3+7+10)sinA

= 20sinA

Subtract: 7sinθ from 18sinθ

Solution:

= 18sinθ − 7sinθ

= (18 − 7)sinθ

= 11sinθ

Multiply: 3tanA × 7tan^{4}A

Solution:

= 3tanA × 7tan^{4}A

= 21tan^{1+4}A

= 21tan^{5}A

Find the product of: (3sinθ + 2cosθ) (3sinθ − 2cosθ)

= (3sinθ + 2cosθ) (3sinθ− 2cosθ)

= (3sinθ)^{2}− (2cosθ)^{2}

= 9sin^{2}θ − 4cos^{2}θ

Divide: 12tan^{7}A by 3tan^{3}A

Solution:

= 12tan^{7}A÷ 3tan^{3}A

= \(\frac{12tan^7A}{3tan^3A}\)

= 4tan^{7−3}A

= 4tan^{4}A

Simplify: sinθ (1 − cosθ) + cosθ (1 + sinθ)

= sinθ(1− cosθ) + cosθ(1 + sinθ)

= sinθ − sinθ.cosθ + cosθ + sinθ.cosθ

= sinθ + cosθ

Simplify: \(\frac{(sinA − cosA)^2}{sin^2A − cos^2A}\)

= \(\frac{(sinA− cosA)^2}{sin^2− cos^2}\)

= \(\frac{(sinA− cosA)(sinA− cosA)}{(sinA + cosA)(sinA− cosA)}\)

= \(\frac{sinA− cosA}{sinA + cosA}\)

Prove that: \(\frac{sinθ.cosecθ}{secθ}\) = cosθ

Solution:

L.H.S. = \(\frac{sinθ.cosecθ}{secθ}\)

= sinθ cosecθ÷ secθ

= sinθ× \(\frac{1}{sinθ}\)÷ \(\frac{1}{cosθ}\)

= sinθ× \(\frac{1}{sinθ}\)× cosθ

= cosθ

= R.H.S. proved.

Proved that: secθ \(\sqrt{1 − cos^2θ}\) = tanθ

Solution:

L.H.S. = secθ \(\sqrt{1− cos^2θ}\)

= \(\frac{1}{cosθ}\) . sinθ

= \(\frac{sinθ}{cosθ}\)

= tanθ

= R.H.S. proved.

Prove that: sin^{2}θ + cos^{2}θ.tan^{2}θ = 2sin^{2}θ

Solution:

L.H.S. = sin^{2}θ + cos^{2}θ.tan^{2}θ

= sin^{2}θ + cos^{2}θ \(\frac{sin^2θ}{cos^2θ}\)

= sin^{2}θ + sin^{2}θ

= 2sin^{2}θ

= R.H.S. proved

Prove that: (sinA + cosA)^{2} + (sinA − cosA)^{2}^{ } = 2

Solution:

L.H.S. = (sinA + cosA)^{2} + (sinA − cosA)^{2}

= sin^{2}A + 2sinA.cosA + cos^{2}A + sin^{2}A − 2sinA.cosA + cos^{2}A

= (sin^{2}A + cos^{2}A) + (sin^{2}A + cos^{2}A)

= 1 + 1

= 2

= R.H.S. proved

Prove that: \(\frac{1}{tanA + cotA}\) = sinA.cosA

Solution:

L.H.S. = \(\frac{1}{tanA + cotA}\)

= \(\frac {1}{\frac {sinA}{cosA} + \frac {cosA}{sinA}}\)

= \(\frac{1}{\frac{sin^2A + cos^2A}{sinA.cosA}}\)

= \(\frac{1}{\frac{1}{sinA.cosA}}\)

= sinA.cosA

= R.H.S. proved

Prove that: \(\frac{sin^4θ − cos^4θ}{sinθ + cosθ}\) = sinθ − cosθ

Solution:

L.H.S. = \(\frac{sin^4θ− cos^4θ}{sinθ + cosθ}\)

= \(\frac{(sin^2θ)^2 − (cos^2θ)^2}{sinθ + cosθ}\)

= \(\frac{(sin^2θ + cos^2θ) (sin^2θ− cos^2θ)}{sinθ + cosθ}\)

= \(\frac{1× (sinθ + cosθ) (sinθ− cosθ)}{sinθ + cosθ}\)

= sinθ− cosθ

= R.H.S. proved

Prove that: \(\frac{cosA}{1 + sinA}\) + \(\frac{cosA}{1 − sinA}\) = 2secA

Solution:

L.H.S. = \(\frac{cosA}{1 + sinA} + \frac{cosA}{1− sinA}\)

= \(\frac{cosA (1− sinA) + cosA (1 + sinA)}{(1 + sinA) (1− sinA)}\)

= \(\frac{cosA− sinA.cosA + sinA.cosA}{1− sin^2A}\)

= \(\frac{2cosA}{cos^2A}\)

= \(\frac{2}{cosA}\)

= 2secA

= R.H.S. proved

Prove that: \(\frac{1 − sin^4A}{cos^4}\) = 1 + 2tan^{2}A

Solution:

L.H.S.

= \(\frac{1− sin^4}{cos4A}\)

= \(\frac{1− sin^4A}{cos^4A}\) − \(\frac{sin^4A}{cos^4A}\)

= sec^{4}A − tan^{4}A

= (sec^{2}A)^{2}− (tan^{2}A)^{2}

= (sec^{2}A− tan^{2}A) (sec^{2}A + tan^{2}A)

= (1 + tan^{2}A + tan^{2}A)

= 1 + 2tan^{2}A

= R.H.S. proved

Prove that: (xcosθ + ysinθ)^{2} + (xsinθ − ycosθ)^{2} = x^{2} + y^{2}

^{ }

Solution:

L.H.S

= (xcosθ + ysinθ)^{2}+ (xsinθ− yccosθ)^{2}

= (xcosθ)^{2}+ 2xysinθ.cosθ + (ysinθ)^{2}+ (xsinθ)^{2}− 2xysinθ.cosθ + (ycosθ)^{2}

= x^{2}cos^{2}θ + y^{2}sin^{2}θ + x^{2}sin^{2}θ+ y^{2}cos^{2}θ

= x^{2}(cos^{2}θ + sin^{2}θ) + y^{2}(sin^{2}θ + cos^{2}θ)

= (x^{2}+ y^{2}) (cos^{2}θ + sin^{2}θ)

= (x^{2}+ y^{2}) \(\times\) 1 [\(\because\) (cos^{2}θ + sin^{2}θ) =1]

= (x^{2}+ y^{2}) R.H.S. proved

Prove that: \(\sqrt\frac{1 + cosθ}{1 – cosθ}\) = cosecθ + cotθ

Solution:

L.H.S. =\(\sqrt\frac{1 + cosθ}{1 – cosθ}\)

= \(\sqrt{\frac{1 + cosθ}{1− cosθ}× \frac{1 + cosθ}{1− cosθ}}\)

= \(\sqrt\frac{(1 + cosθ)^2}{1− cos^2θ}\)

= \(\sqrt\frac{(1 + cosθ)^2}{sin^2θ}\)

= \(\frac{1 + cosθ}{sinθ}\)

= \(\frac{1}{sinθ} + \frac{cosθ}{sinθ}\)

= cosecθ + cotθ

= R.H.S. proved

Express all trigonometric ratios of θ in terms of sinθ.

Solution:

Let, ABC be a right angled triangle where ∠B = 90° and ∠C = θ = k

Let, sinθ = k

Then,

or, sinθ = \(\frac{k}{1}\) = \(\frac{p}{h}\)

∴ p = k, h = 1

Now, from right angled ΔABC,

or, b = \(\sqrt{h^2 − p^2}\)

or, b = \(\sqrt{1 − k^2}\)

or, b = \(\sqrt{1− sin^2θ}\)

Then,

cosecθ = \(\frac{h}{p}\) = \(\frac{1}{k}\) = \(\frac{1}{sinθ}\)

cosθ = \(\frac{b}{h}\) = \(\sqrt\frac{1− sin^2θ}{1}\) = \(\sqrt{1−sin^2θ}\)

secθ = \(\frac{h}{b}\) = \(\frac{1}{\sqrt{1− k^2}}\)

tanθ = \(\frac{p}{b}\) = \(\frac{k}{\sqrt{1− k^2}}\) = \(\frac{sinθ}{\sqrt{1− sin^2θ}}\)

cotθ = \(\frac{b}{p}\) = \(\frac{\sqrt{1− sin^2θ}}{sinθ}\)

Find the values of other trigonometric ratios when cosα = \(\frac{5}{13}\)

Solutions:

Here,

cosα = \(\frac{5}{13}\) = \(\frac{b}{h}\)

Then,

b = 5, h = 13, p= ?

By using Pythagoras theorem, we have,

or, p^{2}= h^{2}− b^{2}

or, p = \(\sqrt{h^2 − b^2}\)

or, p = \(\sqrt{13^2 − 5^2}\)

or, p = \(\sqrt{169 − 25}\)

or, p = \(\sqrt{144}\)

∴ p = 12

Now,

secα = \(\frac{h}{b}\) = \(\frac{13}{5}\)

sinα = \(\frac{p}{h}\) = \(\frac{12}{13}\)

cosecα = \(\frac{h}{p}\) = \(\frac{13}{12}\)

tanα = \(\frac{p}{b}\) = \(\frac{12}{5}\)

cotα = \(\frac{b}{p}\) = \(\frac{5}{12}\)

If sinθ − cosθ = 0, find the values of cosecθ, sinθ and cosθ.

Solution:

Here,

or, sinθ − cosθ = 0

or, sinθ = cosθ

or, tanθ = 1

We have,

tanθ = \(\frac{p}{b}\) = \(\frac{1}{1}\)

Then,

p = 1, b = 1

By Pythagoras theorem,

or, h^{2} = p^{2} + b^{2}

or, h = \(\sqrt{p^2 + b^2}\)

or, h = \(\sqrt{1^2 +1^2}\)

or, h = \(\sqrt{1 + 1}\)

or, h = \(\sqrt{2}\)

∴ h = \(\sqrt{2}\)

Now,

cosecθ = \(\frac{h}{p}\) = \(\frac{\sqrt{2}}{1}\) = \(\sqrt{2}\)

sinθ = \(\frac{p}{h}\) = \(\frac{1}{\sqrt{2}}\)

If sinθ = \(\frac{m}{n}\), proved that \(\sqrt{n^2 − m^2}\) tanθ = m

Solution:

Here,

or, sinθ = \(\frac{m}{n}\) = \(\frac{p}{h}\)

Then,

p = m, h = n

We have,

or, b^{2} = h^{2}− p^{2}

or, b^{2} = n^{2}− m^{2}

or, b = \(\sqrt{n^2 − m^2}\)

Now,

or, tanθ = \(\frac{p}{b}\)

or, tanθ = \(\frac{m}{\sqrt{n^2− m^2}}\)

or, \(\sqrt{n^2− m^2}\) tanθ = m

\(\therefore\) L.H.S = R.H.S proved

If 3cotθ = 4, find the value of \(\frac{3cosθ + 2sinθ}{3cosθ − 2sinθ}\)

Solution:

Here,

or, 3cotθ = 4

We have,

or, cotθ = \(\frac{b}{p}\) = \(\frac{4}{3}\)

Then,

b = 4, p = 3, h = ?

By Pythagoras theorem,

or, h^{2} = p^{2} + b^{2}

or, h^{2} = 3^{2} + 4^{2}

or, h^{2} = 9 + 16

or, h^{2} = \(\sqrt{25}\)

∴ h = 5

Now,

= \(\frac{3cosθ + 2sinθ}{3cosθ− 2sinθ}\)

= {3 \((\frac{b}{h}) + 2 (\frac{p}{h})\)}÷ {3 \((\frac{b}{h})− 2 (\frac{p}{h})\)}

= (3× \(\frac{4}{5}\) + 2× \(\frac{3}{5}\))÷ (3× \(\frac{4}{5}\)− 2× \(\frac{3}{5}\))

= (\(\frac{12}{5} + \frac{6}{5}\))÷ (\(\frac{12}{5}− \frac{6}{5}\))

= \(\frac{18}{5}\)÷ \(\frac{6}{5}\)

= \(\frac{18}{5}\)× \(\frac{5}{6}\)

= 3

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