Question

evaluate cosec^2 63°+tan^2 24°cot^2 66°+sec^2 27°+sin 63°+cos 63°sin27° +sec63°/2(cosec^2 65°-tan^2 25°)

Answer 1

Hello,
(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°);

Being:tan (90-A)= cot A, Cosec (90-A)= secA, cot (90-A) = tanA
then

= (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°);

Being:sin²A+cos²A=1, cosecA= 1/sinA,cosec²- cot²A=1
then

= 1+ {( 1+1)/2×1};

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

Bye :-)

Answer 2

Answer is given below......

(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°);

Being:tan (90-A)= cot A, Cosec (90-A)= secA, cot (90-A) = tanA
then

= (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°);

Being:tan (90-A)= cot A, Cosec (90-A)= secA, cot (90-A) = tanA
then

= (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°);

Being:sin²A+cos²A=1, cosecA= 1/sinA,cosec²- cot²A=1
then

= 1+ {( 1+1)/2×1};

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

Very long answer lol....took very much time to write