evaluate cosec^2 63+tan^2 24/cot^2 66+sec^2 27+sin^2 63+cos 63.sin 27+sin 27.sec63/2(cosec^2 65-tan^2 25)
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Answered by
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(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...
= (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...
Answered by
61
Answer:
(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
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