Math, asked by rahahul2863, 1 year ago

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


Shaik96: Can u specify the brackets correctly?

Answers

Answered by alessre
15
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 :-)
Answered by KnowMore
7
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
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