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INTRODUCTION TO TRIGONOMETRY 191<br />

Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and<br />

sec A are not defined for A = 90°. So, (3) is true for all A such that 0° A ✁ 90°.<br />

Let us see what we get on dividing (1) by BC 2 . We get<br />

AB<br />

BC<br />

2 2<br />

BC<br />

✂ =<br />

BC<br />

2 2<br />

AC<br />

BC<br />

2<br />

2<br />

i.e.,<br />

2 2<br />

AB ✄ BC ☎ ✄ ☎<br />

✆ ✝ ✞ ✝ ✞<br />

BC BC<br />

✟ ✠ ✟ ✠<br />

=<br />

AC ✄ ☎<br />

✝ ✞<br />

BC<br />

✟<br />

2<br />

✠<br />

i.e., cot 2 A + 1 = cosec 2 A (4)<br />

Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for<br />

all A such that 0° < A 90°.<br />

Using these identities, we can express each trigonometric ratio in terms of other<br />

trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the<br />

values of other trigonometric ratios.<br />

tan A =<br />

Let us see how we can do this using these identities. Suppose we know that<br />

1<br />

3 ✡ Then, cot A = 3 .<br />

1 4<br />

Since, sec 2 A = 1 + tan 2 A ☛ = ☞ 1 , sec A =<br />

3 3<br />

23 , and cos A = 3<br />

2 ✡<br />

2 3 1<br />

Again, sin A = 1✌ cos ✍ ✌ ✍ A 1 . Therefore, cosec A = 2.<br />

4 2<br />

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

Solution : Since<br />

cos 2 A + sin 2 A = 1, therefore,<br />

cos 2 A = 1 – sin 2 A, i.e., cos A =<br />

2<br />

1 sin A<br />

✎<br />

✏<br />

2<br />

This gives cos A = 1✏<br />

sin A<br />

(Why?)<br />

Hence,<br />

tan A = sin A<br />

cos A = sin A 1 1<br />

and sec A = ✑<br />

2 cos A<br />

2<br />

1–sin A 1 sin A<br />

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