[Maths Class Notes] on Trigonometric Identities Pdf for Exam

The Six Trigonometric Ratios that are Frequently Used are:

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• Sine

• Cosine

• Tangent

• Secant

• Cosecant

• Cotangent

The sides like – opposite side, adjacent side, and hypotenuse side –  of a right-angled triangle, is used to define the angles that are mentioned above. With the help of the six trigonometric ratios, you can derive all the six identities. In this article, we have discussed the trigonometric identities list that is going to come in handy while you solve the problems. In addition to that, we also have shown you how some of these identities are derived and also solved some problems for your better understanding. 

List of Trigonometric Identities

There are a lot of trigonometric identities that are used to solve Trigonometric Identities problems based on trigonometric equations. Find all the trigonometric identities and equations below:

Reciprocal Identities

• sin k = 1 / cosec k or cosec k = 1 / sin kt

• cos k = 1 / sec k or sec k = 1 / cos kt

• tan k = 1 / cot k or cot k = 1 / tan kt

The Pythagorean Identities

• sin2 kt + cos2 kt = 1

• 1 + tan2 kt = sec2 kt

• Cosec2 kt = 1 + cot2 kt

The Ratio Identities

• tan kt = sin kt / cos kt

• cot kt = cos kt / sin kt

The Opposite Angle Identities

• sin ( – kt  ) = – sin kt

• cos ( – kt  ) = cos kt

• tan ( – kt ) = – tan kt

• cot ( – kt  ) = – cot kt

• sec ( – kt  ) = sec kt

• cosec ( – kt  ) = – cosec kt

The Complementary Angle Identities

The Angle’s Sum and Difference identities

Let us consider two angles – p and q.  Now, the trigonometric sum and difference identities are:

• sin ( p + q ) = sin ( p ) * cos ( q ) + cos ( p ) * sin ( q )

• sin ( p – q ) = sin ( p ) * cos ( q ) – cos ( p ) * sin ( q )

• cos ( p + q ) = cos ( p ) * cos ( q ) – sin ( p ) * sin ( q )

• cos ( p + q ) = cos ( p ) * cos ( q ) + sin ( p ) * sin ( q )

• tan ( p + q ) = [frac{text{tanp + tanq}}{text{1 – tanp * tanq}}] 

• tan ( p + q ) = [frac{text{tanp – tanq}}{text{1 + tanp * tanq}}]

The Trigonometric Identities Formula

The trigonometric identity is represented with the help of an equation that has trigonometric ratios. Here, we shall understand the basics of trigonometric identities and their proofs.

Consider a triangle PQR. This triangle is right-angled at the point Q. 

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When you apply the Pythagoras theorem to the above triangle, you get:

RP2 = QR2 + PQ2  . . . . . . . . . . . . . . . . .. . . . . . . . . Equation ( 1 )

( hypotenuse )2  = ( base )2 + ( adjacent )2

Now, let us prove the most commonly used trigonometric identities.

Trigonometric Identity 1

First, divide each term with RP2. We get:

[frac{PR^{2}}{PR^{2}}] = [frac{PQ^{2}}{PR^{2}}] + [frac{QR^{2}}{PR^{2}}]

⇒ [frac{PQ^{2}}{PR^{2}}] + [frac{QR^{2}}{PR^{2}}] = 1

⇒ [(frac{PQ}{PR})^{2}] + [(frac{QR}{PR})^{2}] = 1. . . . . . . . . . . . . . . . . Equation ( 2 )

Trigonometric Identity 2

First, divide each term with RP2. We get:

[frac{PR^{2}}{PQ^{2}}] = [frac{PQ^{2}}{PQ^{2}}] + [frac{QR^{2}}{PQ^{2}}]

⇒ [frac{PR^{2}}{PQ^{2}}] = 1 + [frac{QR^{2}}{PQ^{2}}]

⇒ [(frac{PR}{PQ})^{2}] = 1 + [(frac{QR}{PQ})^{2}]. . . . . . . . . . . . . . . . .. . . . . . . . . Equation ( 3 )

With the help of the trigonometric ratio, we can find that:

[frac{PR}{PQ}] = [frac{hypotenuse}{adjacent}] = secant kt 

In a similar manner, 

[frac{QR}{PQ}] = [frac{opposite}{adjacent}] = tan kt

Now, replace the value of [frac{PR}{QR}]and [frac{QR}{PQ}] in Eqn. 3, we get:

1 + tan2 k = sec2 kt

The identity that is obtained above stands true for 0 ≤ k ≤ 90° only since tan kt is not defined. 

Trigonometric Identity 3

First, divide each term with BC2. We get:

[frac{PR^{2}}{QR^{2}}] = [frac{PQ^{2}}{QR^{2}}] + [frac{QR^{2}}{QR^{2}}]

⇒ [frac{PR^{2}}{QR^{2}}] = 1 + [frac{PQ^{2}}{QR^{2}}]

⇒ [(frac{PR}{QR})^{2}] = 1 + [(frac{PQ}{QR})^{2}]. . . . . . . . . . . . . . . . .. . . . . . . . . Equation ( 4 )

With the help of the trigonometric ratio, we can find that:

[frac{PR}{PQ}] = [frac{hypotenuse}{opposite}] = cosec kt

In a similar manner, 

[frac{QR}{PQ}] = [frac{adjacent}{opposite}] = cot kt

Now, replace the value of [frac{PR}{PQ}] and [frac{QR}{PQ}] in Eqn. 4, we get:

cosec2 kt = 1 + cot2 kt