Question

1.sin66°-cos24°=0
2.cos257°+cos233°=1​

Answer 1

Answer:

Given, sin66° - cos24° = 0

Need to prove the given equation as zero

⇒ we know that cos(90 - θ) = sinθ

∴ cos24° = cos(90 - 66)°

= sin66° - - - eq (1)

⇒ sin66° - cos24° = 0

[substitute eq(1)]

⇒ sin66° - sin66° = 0

Hence, LHS = RHS

Answer 2

EXPLANATION.

(1) = sin66° - cos24° = 0.

As we know that,

we can write cos 24° as,

⇒ cos24° = cos(90° - 66°).

⇒ sin66° - cos(90° - 66°).

⇒ sin66° - sin66°

⇒ 0.

HENCE PROVED.

(2) = cos²57° + cos²33° = 1.

As we know that,

we can write cos²33° as,

⇒ cos²33° = cos²(90° - 57°).

⇒ cos²57° + cos²(90° - 57°).

⇒ cos²57° + sin²57°.

⇒ 1.

HENCE PROVED.

                                                                                                                                         

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²∅ + cos²∅ = 1.

(2) = 1 + tan²∅ = sec²∅.

(3) = 1 + cot²∅ = cosec²∅.

Sign of trigonometric ratios of functions.

(1) = in first quadrant all are positive.

(2) = in second quadrant only [sin∅ and cosec∅] are positive.

(3) = in third quadrant [ tan∅ and cot∅ ] are positive.

(4) = in fourth quadrant [ cos∅ and sec∅ ] are positive.