if 450°< theta <540° and sin theta =12/13. then calculate sin theta÷2 and cos theta ÷2
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Answers
Answer:
Valueof
2sinθcosθ
(sin
2
θ−cos
2
θ)
×
tan
2
θ
1
=
3456
595
Step-by-step explanation:
\begin{gathered}Given ,\\sin\theta = \frac{12}{13}--(1)\end{gathered}
Given,
sinθ=
13
12
−−(1)
\begin{gathered}cos^{2}\theta = 1-sin^{2}\theta \\=1-(\frac{12}{13})^{2}\\=1-\frac{144}{169}\\=\frac{169-144}{169}\\=\frac{25}{169}\end{gathered}
cos
2
θ=1−sin
2
θ
=1−(
13
12
)
2
=1−
169
144
=
169
169−144
=
169
25
\begin{gathered}cos\theta = \sqrt{\frac{25}{169}}\\=\frac{5}{13}--(2)\end{gathered}
cosθ=
169
25
=
13
5
−−(2)
\begin{gathered}tan\theta = \frac{sin\theta}{cos\theta}\\=\frac{\frac{12}{13}}{\frac{5}{13}}\\=\frac{12}{5}--(3)\end{gathered}
tanθ=
cosθ
sinθ
=
13
5
13
12
=
5
12
−−(3)
\begin{gathered}Now,Value \: of \\ \frac{(sin^{2}\theta-cos^{2}\theta)}{2sin\theta cos\theta}\times \frac{1}{tan^{2}\theta}\end{gathered}
Now,Valueof
2sinθcosθ
(sin
2
θ−cos
2
θ)
×
tan
2
θ
1
=\frac{sin^{2}\theta-(1-sin^{2}\theta)}{2sin\theta cos\theta }\times \frac{1}{tan^{2}\theta}=
2sinθcosθ
sin
2
θ−(1−sin
2
θ)
×
tan
2
θ
1
=\frac{(2sin^{2}\theta-1)}{2sin\theta cos\theta}\times \frac{1}{tan^{2}\theta}=
2sinθcosθ
(2sin
2
θ−1)
×
tan
2
θ
1
=\frac{[2\big(\frac{12}{13}\big)^{2}-1]}{2\times \frac{12}{13}}\times \frac{5}{13}\times \frac{1}{\big(\frac{12}{5}\big)^{2}}=
2×
13
12
[2(
13
12
)
2
−1]
×
13
5
×
(
5
12
)
2
1
=\frac{2\times \frac{144}{169}-1}{2\times \frac{12}{13}\times \frac{5}{13}}\times \frac{25}{144}=
2×
13
12
×
13
5
2×
169
144
−1
×
144
25
=\frac{\frac{(288-169)}{169}}{\frac{120}{169}}\times \frac{25}{144}=
169
120
169
(288−169)
×
144
25
=\frac{119\times 5}{24 \times 144}=\frac{ 595}{3456}=
24×144
119×5
=
3456
595