If CosecA - SinA = a3, SecA - CosA = b3 , then Prove That a2b2 (a2 + b2) = - Maths - Trigonometric Functions

Question

If CosecA - SinA = a3, SecA - CosA = b3 ,
then Prove That a2b2 (a2 +
b2) = 1
If Tn = SinnA + CosnA, P T ( i
) : T3 - T5 / T1 = T5 -
T7 / T3 ; ( ii ) : 2T6 -
3T4 + 1 = 0 ; ( iii ) : 6T10 -
15T8 + 10T6 - 1 = 0

Simran Preet Arora · Answer

cosecA-sinA=a3
=> 1/sinA-sinA=a3
=>(1-sin2A)/sinA=a3
=>cos2A/sinA=a3 ----- i
Again, secA- CosA= b3
=>1/cosA-CosA=b3
=>(1-cos2A)cosA=b3
=> sin2A/cosA=b3 ------ ii
Now, i x ii =>
cos2A/sinA x sin2A/ cosA =
a3b3
=>cosAsinA=a3b3------ iii
i --> cos(square)A/sinA=a(cube)
=>cos2A=sinA
a3
=>cos3A=cosAsinA a3 [multiplying both
sides by 'cosA']
=>cos3A= a3b3 a3
[from iii]
=>cos3A=A6b3
=>CosA=a2b
=>cos2A=a4b2---- iv
ii -->
sin2A/cosA=b3
=>sin2A=cosA b3
=>sin2A sinA=cosA sinA b3 [multiplying
both sides by 'sinA']
=>sin3A=cosAsinAb3
=>sin3A=a3b6
=>sinA=ab2
=>sin2A=a2b4 ------ v
Now, iv + v =>

sin2A+cos2A=a2b4+a4b2

=>1=a2b2(a2+b2)
Proved

If CosecA - SinA = a3, SecA - CosA = b3 , then Prove That a2b2 (a2 + b2) = 1 If Tn = SinnA + CosnA, P.T ( i ) : T3 - T5 / T1 = T5 - T7 / T3 ; ( ii ) : 2T6 - 3T4 + 1 = 0 ; ( iii ) : 6T10 - 15T8 + 10T6 - 1 = 0

If CosecA - SinA = a³, SecA - CosA = b³ , then Prove That a²b² (a²+ b²) = 1

If T_n = SinⁿA + CosⁿA, P.T ( i ) : T₃ - T₅ / T₁ = T₅ - T₇ / T₃ ; ( ii ) : 2T₆ - 3T₄ + 1 = 0 ; ( iii ) : 6T₁₀ - 15T₈ + 10T₆ - 1 = 0