NCERT Class 10 Mathematics

194 MATHEMATICS

(iii)

tan cot

1sec cosec

1cot 1 tan

✁ ✂ ✁

✄ ✄

[Hint : Write the expression in terms of sin ☎ and cos ☎]

(iv)

1secA sinA^2

sec A 1 – cos A

✆ ✝

[Hint : Simplify LHS and RHS separately]

(v) cos A – sin A + 1 cosec A + cot A,

cos A + sin A – 1

✂ using the identity cosec^2 A = 1 + cot^2 A.

(vi)

1sinA

sec A + tan A

1 – sin A

✞

✟ (vii)

3

3

sin 2 sin tan

2cos cos

✠✡ ✠

☛ ✠

✠✡ ✠

(viii) (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A

(ix)

(cosec A – sin A) (sec A – cos A)^1

tan A + cot A

✂

[Hint : Simplify LHS and RHS separately]

(x)

2 2

2

1tanA 1tanA

1+cot A 1 – cot A

☞ ✍ ✌ ☞ ✎ ✌

✑ ✒✏✑ ✒

✓ ✔ ✓ ✔

= tan^2 A

8.6 Summary

In this chapter, you have studied the following points :

1. In a right triangle ABC, right-angled at B,

sin A =

side opposite to angle A, side adjacent to angle A

cos A =

hypotenuse hypotenuse

tan A =

side opposite to angle A

side adjacent to angle A.

2.

111s, inA

cosec A = ; sec A = ; tan A = tan A =

sin A cos A cot A cos A

.

3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric
   ratios of the angle can be easily determined.


4. The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.


5. The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is
   always greater than or equal to 1.


6. sin (90° – A) = cos A, cos (90° – A) = sin A;
   tan (90° – A) = cot A, cot (90° – A) = tan A;
   sec (90° – A) = cosec A, cosec (90° – A) = sec A.


7. sin^2 A + cos^2 A = 1,
   sec^2 A – tan^2 A = 1 for 0° ✕ A < 90°,
   cosec^2 A = 1 + cot^2 A for 0° < A ✕ 90º.