Math, asked by prtshpndy, 6 months ago

sin-63° +sin227°
sec248°-cot242
(tan? (6 + 45°) + cosec(45° – 0)],
then coso equals [ 0º = 0 < 45°)
1
3​

Answers

Answered by itsbiswaa
1

Answer:

Answer Expert Verified

4.0/5

425

nikitasingh79

Genius

15.5K answers

34.9M people helped

(cosec²63°+ tan² 24°/ cot²66° +sec²27°) + (sin²63°+ cos63° sin27° + sin27ºsec63° /2(Cosec ²65° - tan²65°- tan²25°)

= (cosec²63°+ tan² 24°/ tan²(90°-66) +cosec²(90°- 27°) + (sin²63°+ cos63° cos(90°-27°) + sin27ºcosec(90°- 63°) /2(Cosec ²65° - cot²(90°-25°)

[ tan (90-A)= cot A, Cosec (90-A)= secA, cot (90-A= tanA]

= (cosec²63°+ tan² 24°/ tan²24° +cosec²63°) + (sin²63°+ cos²63° + sin27ºcosec27° /2(Cosec ²65° - cot²65°)

= (cosec²63°+ tan² 24°/ tan²24° +cosec²63°) + (sin²63°+ cos²63° + sin27º×1/sin27° /2(Cosec ²65° - cot²65°)

[sin²A+cos²A=1, cosecA= 1/sinA,cosec²- cot²A=1]

= 1+ {( 1+1)/2×1}

= 1+2/2= 1+ 1= 2

Hope this will help you...

plz mark as brainliast answer✔✔✔✔✔✔✔✔✔✔✔✔

thnks my answers also❤❤❤❤❤

Step-by-step explanation:

Similar questions