For any θ∈(π/4,π/2) the expression 3(sin⁡〖θ-cos⁡θ 〗 )^4+6(sin⁡〖θ+cos⁡θ 〗 )^2+4sin^6⁡θ equals

For any θ∈(π/4,π/2) the expression 3(sin⁡〖θ-cos⁡θ 〗 )^4+6(sin⁡〖θ+cos⁡θ 〗 )^2+4sin^6⁡θ equals:
(a) 13-4 cos^2⁡〖θ+6 sin^2⁡〖θ cos^2⁡θ 〗 〗
(b) 13-4 cos^6⁡θ
(c) 13-4 cos^2⁡〖θ+6 cos^4⁡θ 〗
(d) 13-4 cos^4⁡〖θ+2 sin^2⁡〖θ cos^2⁡θ 〗 〗
Ans: b
Sol.
3(sin⁡〖θ-cos⁡θ 〗 )^4+6(sin⁡〖θ+cos⁡θ 〗 )^2+4sin^6⁡θ
=3(1-2 sin⁡〖θ cos⁡θ 〗 )^2+6(1+2 sin⁡〖θ cos⁡θ 〗 )+4sin^6⁡θ
=3(1+4 sin^2⁡cos^2⁡〖θ-4 sin⁡〖θ cos⁡θ 〗 〗 )+6+12sin⁡〖θ cos⁡〖θ+4 sin^6⁡θ 〗 〗
=9+12sin^2⁡〖θ+cos^2⁡〖θ+4 sin^6⁡θ 〗 〗
=9+12cos^2⁡θ(1-cos^2⁡θ )+4(1-cos^2⁡θ )^3
=9+12cos^2⁡〖θ-12 cos^4⁡〖θ+4(1-cos^6⁡〖θ-3 cos^2⁡〖θ+3 cos^4⁡θ 〗 〗 )〗 〗
=9+4-4cos^6⁡θ
=13-4cos^6⁡θ

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Видео For any θ∈(π/4,π/2) the expression 3(sin⁡〖θ-cos⁡θ 〗 )^4+6(sin⁡〖θ+cos⁡θ 〗 )^2+4sin^6⁡θ equals канала Math Master - Impetus Gurukul