If root3tantheeta=3sintheeta find the value of sin squaretheeta-cos squaretheeta
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√3tanФ=3sinФ (theeta≡Ф)
or, √3 sinФ/cosФ=3sinФ
or, 1/cosФ=3sinФ/√3sinФ
or, 1/cosФ=√3
or, cosФ=1/√3
∴,sinФ=√1-cos²Ф=√1-1/3=√(3-1)/3=√2/3; [∵, sin²Ф+cos²Ф=1]
∴, sin²Ф-cos²Ф=(√2/3)²-(1/√3)²=2/3-1/3=(2-1)/3=1/3
or, √3 sinФ/cosФ=3sinФ
or, 1/cosФ=3sinФ/√3sinФ
or, 1/cosФ=√3
or, cosФ=1/√3
∴,sinФ=√1-cos²Ф=√1-1/3=√(3-1)/3=√2/3; [∵, sin²Ф+cos²Ф=1]
∴, sin²Ф-cos²Ф=(√2/3)²-(1/√3)²=2/3-1/3=(2-1)/3=1/3
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