Question
Download Solution PDFIf \(\frac{\sin θ-2 \sin ^{3} θ}{2 \cos ^{3} θ-\cos θ}=\frac{1}{\sqrt{7}}\), 0° < θ < 90°, then what is the value of (sec2 θ + cosec2 θ) ÷ (sec2 θ - cosec2 θ)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
\(\frac{\sin θ-2 \sin ^{3} θ}{2 \cos ^{3} θ-\cos θ}=\frac{1}{\sqrt{7}}\), 0° < θ < 90°
Formula used:
tan θ = sin θ / cos θ
sin2 θ + cos2 θ = 1
sec2 θ = 1 + tan2 θ
cosec2 θ = 1 + cot2 θ = 1 + 1/tan2 θ
Calculation:
\(\frac{\sin θ(1-2 \sin ^{2} θ)}{\cos θ(2 \cos ^{2} θ-1)}=\frac{1}{\sqrt{7}}\)
\(\frac{\sin θ(1-2(1-\cos^2 θ))}{\cos θ(2 \cos ^{2} θ-1)}=\frac{1}{\sqrt{7}}\)
\(\frac{\sin θ(2\cos^2 θ-1)}{\cos θ(2 \cos ^{2} θ-1)}=\frac{1}{\sqrt{7}}\)
\(\frac{\sin θ}{\cos θ}=\frac{1}{\sqrt{7}}\)
tan θ = 1/√7
sec2 θ = 1 + tan2 θ = 1 + (1/7) = 8/7
cosec2 θ = 1 + cot2 θ = 1 + 7 = 8
Now, substitute into the expression:
(sec2 θ + cosec2 θ) / (sec2 θ - cosec2 θ)
= (8/7 + 8) / (8/7 - 8)
= (8 + 56) / (8 - 56)
= 64 / (-48)
= -64 / 48
= -4 / 3 = \(-1 \frac{1}{3}\)
∴ The value of (sec2 θ + cosec2 θ) / (sec2 θ - cosec2 θ) is \(-1 \frac{1}{3}\).
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