Math, asked by Arpita102028, 2 months ago

1) Let us show that
a) sin66° - cos24° = 0
b) cos²57° + cos²33° = 1
c) cos²75° - sin²15° = 0
d) cosec²48° - tan²42° = 1
e) sec70°sin20°+cos20°cosec70° = 2​

Answers

Answered by misscutie94
60

Answer:

Question :-

a) sin66° - cos24° = 0

b) cos²57 + cos²33° = 1

c) cos²75° - sin²15° = 0

d) cosec²48° - tan²42° = 1

e) sec70° sin20° + cos20° cosec70° = 2

Solution :-

a) sin66° - cos24° = 0

L.H.S.

= sin66° - cos24°

= sin(90° - 24°) - cos24°

= cos24° - cos24°

= 0

∴ L.H.S. = R.H.S.

b) cos²57° + cos²33 = 1

L.H.S.

= cos²57° + cos²33°

= cos²(90° - 33°) + cos²33°

= sin²33° + cos²33°

= 1

∴ L.H.S. = R.H.S.

c) cos²75° - sin²15° = 0

L.H.S.

= cos²75° - sin²15°

= cos²(90° - 15°) - sin²15°

= sin²15° - sin²15°

= 0

∴ L.H.S. = R.H.S.

d) cosec²48° - tan²42° = 1

L.H.S.

= cosec²48° - tan²42°

= cosec²(90° - 42°) - tan²42°

= sec²42° - tan²42°

= 1

∴ L.H.S. = R.H.S.

e) sec70° sin20° + cos20° cosec70° = 2

L.H.S.

= sec70° sin20° + cos20° cosec70°

= sec(90° - 20°) sin20° + cos20° cosec(90° - 20°)

= cosec20° . sin20° + cos20° . sec20°

= 1 + 1

= 2

∴ L.H.S. = R.H.S.


amitkumar44481: Great :-)
Answered by amansharma264
33

EXPLANATION.

(1) = sin66° - cos24° = 1.

As we know that,

⇒ sin(90° - 24°) - cos24°.

⇒ cos24° - cos24° = 1.

(2) = cos²57° + cos²33° = 1.

As we know that,

⇒ cos²57° + cos²(90° - 57°).

⇒ cos²57° + sin²57° = 1.

(3) = cos²75° - sin²15° = 0.

As we know that,

⇒ cos²75° - sin²(90° - 75°).

⇒ cos²75° - cos²75° = 0.

(4) = cosec²48° - tan²42° = 1.

⇒ cosec²(90° - 42°) - tan²42°.

⇒ sec²42° - tan²42° = 1.

(5) = sec70°.sin20° + cos20°.cosec70° = 2.

As we know that,

⇒ sec(90° - 20°).sin20° + cos20°.cosec(90° - 20°).

⇒ cosec20°/cosec20° + 1/sec20°.sec20°.

⇒ 1 + 1 = 2.

Similar questions