8.5 Trigonometric Identities

an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.

prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

In ∆ ABC, right-angled at B (see Fig. 8.22), we have:

$$AB^{2} + BC^{2} = AC^{2}$$………… (1)

Dividing each term of (1) by $$AC^{2}$$, we get

$$\frac{AB^{2}}{AC^{2}}+\frac{BC^{2}}{AC^{2}}=\frac{AC^{2}}{AC^{2}}$$

i.e.,  $$(\frac{AB}{AC})^{2}+(\frac{BC}{AC})^{2}=(\frac{AC}{AC})^{2}$$

i.e., $$(cos A)^{2} + (sin A)^{2} = 1$$

i.e., $$cos^{2} A + sin^{2} A = 1$$………………. (2)

This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Let us now divide (1) by $$AB^{2}$$. We get

$$\frac{AB^{2}}{AB^{2}}+\frac{BC^{2}}{AB^{2}}=\frac{AC^{2}}{AB^{2}}$$

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

i.e., $$1 + tan^{2} A = sec^{2} A$$ ……………..(3)

tan A and sec A are not defined for A = 90°. So, (3) is true for all A such that 0° ≤ A < 90°.

Let us see what we get on dividing (1) by $$BC^{2}$$. We get

$$\frac{AB^{2}}{BC^{2}}+\frac{BC^{2}}{BC^{2}}=\frac{AC^{2}}{BC^{2}}$$

i.e.,      $$\frac{AB}{BC}^{2}+\frac{BC}{BC}^{2}=\frac{AC}{BC}$$

i.e.,      $$cot^{2} A + 1 = cosec^{2} A$$ ……….(4)

Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for all A such that 0° < A ≤ 90°.

Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios,

i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Let us see how we can do this using these identities. Suppose we know that

$$tan A=\frac{1}{\sqrt{3}}$$.  Then , $$cot A=\sqrt{3}$$.

Since, $$sec^{2} A=1+tan^{2}A=1+\frac{1}{3}=\frac{4}{3}, sec A=\frac{2}{\sqrt{3}}, \:\:and\:\: cos A=\frac{\sqrt{3}}{2}$$.

Again , $$sin A=\sqrt{1-cos^{2}A}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$$. Therefore , cosec A=2.

Example 12 : Express the ratios cos A, tan A and sec A in terms of sin A.

Solution : Since $$cos^{2} A + sin^{2} A = 1$$, therefore,

$$Cos^{2}A=1-sin^{2}A, i.e., cos A=±\sqrt{1-sin^{2}A}$$

This gives    $$cos A=\sqrt{1-sin^{2}A}$$

Hence,  $$tan A=\frac{sin A}{cos A}=\frac{sin A}{\sqrt{1-sin^{2}}A}$$ and sec $$A=\frac{1}{cos A}=\frac{1}{\sqrt{1-sin^{2}A}}$$

Example 13 : Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Solution :

LHS=$$sec A (1-sin A)(sec A + tan A)=(\frac{1}{cos A})(1-sin A)(\frac{1}{cos A}+\frac{sin A}{cos A})$$

=$$\frac{(1-sin A)(1+sin A)}{cos^{2}A}=\frac{1-sin^{2}A}{cos^{2}A}$$

=$$\frac{cos^{2}A}{cos^{2}A}=1$$=RHS

Example 14 : Prove that $$\frac{cot A-cos A}{cot A + cos A}=\frac{cosec A-1}{cosec A+1}$$

Solution :LHS=$$\frac{cot A-cos A}{cot A + cos A}=\frac{\frac{cos A}{sin A}-cos A}{\frac{cos A}{sin A}+cos A}$$

=$$\frac{cos A (\frac{1}{sin A}-1)}{cos A (\frac{1}{sin A}+1)}=\frac{(\frac{1}{sin A}-1)}{(\frac{1}{(sin A)+1})}=\frac{cosec A-1}{cosec A+1}=RHS$$

Example 15 : Prove that $$\frac{sin\theta – cos \theta +1}{sin\theta + cos \theta -1}=\frac{1}{sec \theta – tan \theta}$$, using the identity $$Sec^{2} θ = 1 + tan^{2} θ$$.

Solution : Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.

LHS=$$\frac{sin θ – cos θ +1}{ sin θ + cos θ -1}=\frac{tan θ -1+ sec θ }{tan θ +1-sec θ}$$

=$$\frac{(tan θ + sec θ)-1 }{(tan θ -sec θ)+1}=\frac{(tan θ + sec θ)-1(tan θ -sec θ) }{(tan θ -sec θ)+1(tan θ -sec θ)}$$

=$$\frac{(tan^{2} θ -sec^{2} θ)-(tan θ-sec θ)}{{tan θ -sec θ +1}(tan θ – sec θ)}$$

=$$\frac{-1-tan θ +sec θ }{(tan θ-sec θ+1)(tan θ -sec θ)}$$

=$$\frac{-1}{tan θ -sec θ }=\frac{1}{sec θ – tan θ }$$

which is the RHS of the identity, we are required to prove.