An identity is an equality which remains true for entire values of the variables involved in the equation. The algebraic identity \(\mathbf{(a+b)^2 = a^2~+~2ab~+~b^2}\) is true for all real numbers. Similarly the identity \((a+b)^3\) = \(a^3~+~b^3~+~3ab(a~+~b)\) is true for every real value of \(a\) and \(b\).

Thus, for an identity \(X\) = \(Y\); where both \(X\) and \(Y\) consists of some variables, the relation \(X\) = \(Y\) holds true. If \(X\) and \(Y\) define the same function, then \(X\) = \(Y\) is an identity.

**Trigonometric Identities**

Similarly, any equation which involves trigonometric ratios of an angle represents a trigonometric identity.

The upcoming discussion covers the fundamental trigonometric identities and their proofs.

Consider a right angle \(∆ABC\) which is right-angled at \(B\) as shown in the given figure.

Applying Pythagoras Theorem in \(∆ABC\);

\((hypotenuse)^2\) = \((perpendicular)^2~+~(base)^2\)

\(AC^2 = AB^2~+~BC^2 ~~~~~~~~~~~~~~~~~~~—(1)\)

Now, divide each term of equation (1) by \(AC^2\), we have

\(\frac{AC^2}{AC^2}\) = \(\frac{AB^2}{AC^2}~+~\frac{BC^2}{AC^2}\)

\(⇒\frac{AB^2}{AC^2}~+~\frac{BC^2}{AC^2}\) = \(1\)

\(⇒(\frac{AB}{AC})^2~+~(\frac{BC}{AC})^2\) = \(1~~~~~~~~~~~~~~~—(2)\)

Making use of trigonometric ratios; for \(∆ABC\) given,

\(\frac{AB}{AC}\) = \(\frac{side~ adjacent~ to~ angle~ A}{hypotenuse}\) = \(cos~A\)

Similarly,

\(\frac{BC}{AC}\) = \(\frac{side ~opposite~ to~ angle~ A}{hypotenuse}\) = \(sin~A\)

Replacing the values of \(\frac{AB}{AC}\) and \(\frac{BC}{AC}\) in the equation (2) with \(cos~A\) and \(sin~A\) respectively gives ,

\(\mathbf{sin^2~A~+~cos^2~A = 1}\)** (IDENTITY 1)
**

This is valid for all the values of \(∠A\) where \(0° ≤ A ≤ 90°\). Hence, this represents a trigonometric identity.

Now Dividing the equation (1) by \(AB^2\), we get

\(\frac{AC^2}{AB^2}\) = \(\frac{AB^2}{AB^2}~+~\frac{BC^2}{AB^2}\)

\(⇒\frac{AC^2}{AB^2}\) = \(1~+~\frac{BC^2}{AB^2} ⇒ (\frac{AC}{AB})^2\) = \(1~+~(\frac{BC}{AB})^2~~~~~~~~~~~~~~~~~~—(3)\)

By referring trigonometric ratios, it can be seen that:

\(\frac{AC}{AB}\) = \(\frac{hypotenuse}{side~ adjacent~ to ~angle~ A}\) = \(sec~A\)

Similarly,

\(\frac{BC}{AB}\) = \(\frac{side~ opposite~ to~ angle~ A}{side ~adjacent~ to~ angle~ A}\) = \(tan~A\)

Replacing the values of \(\frac{AC}{AB}\) and \(\frac{BC}{AB}\) in the equation (3) gives,

\(\mathbf{1~+~tan^2~A = sec^2~A}\) **(****IDENTITY 2)**

As it is known that \(tan~A\) is not defined for \(A\) = 90° therefore the identity obtained above is true for 0 ≤ A <90.

Dividing the equation (1) by \(BC^2\), we get

\(\frac{AC^2}{BC^2}\) = \(\frac{AB^2}{BC^2}~ +~\frac{BC^2}{BC^2}\)

\(⇒\frac{AC^2}{BC^2}\) = \(\frac{AB^2}{BC^2}~+~1\)

\(⇒(\frac{AC}{BC})^2\) = \((\frac{AB}{BC})^2~+~1~~~~~~~~~~~~~~~~~~—(4)\)

By referring trigonometric ratios, it can be seen that:

\(\frac{AC}{BC}\) = \(\frac{hypotenuse}{side ~opposite~ to~ angle~ A}\) = \(cosec~A\)

Also,

\(\frac{AB}{BC}\) = \(\frac{side~ adjacent~ to~ angle~ A}{side~ opposite ~to ~angle~ A}\) = \(cot^2~A\)

Replacing the values of \(\frac{AC}{BC}\) and \(\frac{AB}{BC}\) in the equation (4) gives,

\(\mathbf{cosec~^2~A = 1~+~cot^2~A}\) **(****IDENTITY 3)**

Since \(cosec~A\) and \(cot~A\) are not defined for \(A\) = 0° therefore the identity is obtained is true for all the values of \(A\) except at \(A\) = 0°. Therefore, the identity is true for all \(A\) such that, 0° < \(A\) ≤ 90°.

Thus, the three fundamental trigonometric identities can be listed as:

- \(sin^2~A~+~cos^2~A\) = \(1\)
- \(1~+~tan^2~A\) = \(sec^2~A\)
- \(1~+~cot^2~A\) = \(cosec^2~A\)

**Example:** **Consider a \(∆ABC\), right angled at \(B\). The length of base, \(AB\) = 4 cm and length of perpendicular \(BC\) =3 cm. Find the value of \(sec~A\).**

As the length of perpendicular and base is given; it can be concluded that,

\(tan~A\) = \(\frac{3}{4}\)

Now, using the trigonometric identity **\(1~+~tan^2~A\) = \(sec^2~A\)**

\(sec^2~A\) = \(1~+~(\frac{3}{4})^2\)

\(sec^2~A\) = \(\frac{25}{16}\)

\(⇒sec~A\) = \(±\frac{5}{4}\)

Since, the ratio of lengths is positive, we can neglect \(sec~A\) = \(-\frac{5}{4}\).

Therefore, \(sec~A\) = \(\frac{5}{4}\)

Using these identities, we can solve various mathematical problems. All you need to know about trigonometry and its applications is just a click away. Visit BYJU’S to learn more.