if tan theta=1/root 7 show that 4/3
Answers
Answer:
The value will be \frac { 3 }{ 4 }
4
3
Given:
\tan \theta=\frac{1}{\sqrt({7} )}tanθ=
(
7)
1
To find:
The value of \frac{\csc^{2} \theta-\sec ^{2} \theta}{\csc ^{2} \theta+\sec ^{2} \theta}
csc
2
θ+sec
2
θ
csc
2
θ−sec
2
θ
Solution:
\tan \theta = \frac {\text (opposite side)}{\text (adjacent side)}tanθ=
(adjacentside)
(oppositeside)
\tan \theta =\frac{1}{\sqrt({7} )}tanθ=
(
7)
1
Therefore,
Opposite side = 1 and Adjacent side = \sqrt({7})=
(
7)
By Pythogoras theorem
Hypotenuse ^2 = Adjacent^2+Opposite^2Hypotenuse
2
=Adjacent
2
+Opposite
2
Hypotenuse^2=({\sqrt7})^2 + 1^2Hypotenuse
2
=(
7
)
2
+1
2
Hypotenuse^2 = 7 + 1Hypotenuse
2
=7+1
Hypotenuse = \sqrt 8Hypotenuse=
8
\Sin \theta = \frac {\text(Opposite)}{\text(Hypotenuse)}\Sinθ=
(Hypotenuse)
(Opposite)
Therefore,
\sin \theta=\frac{1}{\sqrt{8}}sinθ=
8
1
\csc \theta=\frac{\sqrt{8}}{1}cscθ=
1
8
\cos \theta = \frac {\text (Adjacent)}{\text(Hypotenuse)}cosθ=
(Hypotenuse)
(Adjacent)
\cos \theta=\frac{\sqrt{7}}{\sqrt{8}}cosθ=
8
7
Therefore,
\sec \theta=\frac{\sqrt{8}}{\sqrt{7}}secθ=
7
8
Now,
\frac{\csc ^{2} \theta-\sec ^{2} \theta}{\csc ^{2} \theta+\sec ^{2} \theta}=
csc
2
θ+sec
2
θ
csc
2
θ−sec
2
θ
=
\frac {\frac{\sqrt{8}^2}{1^2}-\frac{\sqrt{8}^{2}}{\sqrt{7}^{2}}} {\frac{\sqrt{8}^{2}}{1^2}+\frac{\sqrt{8}^{2}}{\sqrt{7}^{2}}}
1
2
8
2
+
7
2
8
2
1
2
8
2
−
7
2
8
2
\frac{8-\frac{8}{7}}{8+\frac{8}{7}}
8+
7
8
8−
7
8
\frac{\frac{56-8}{7}}{\frac{56+8}{7}}
7
56+8
7
56−8
\frac{48}{7} \times \frac{7}{64}
7
48
×
64
7
\frac {48}{64}
64
48
\frac { 3 }{ 4 }
4
3
"