If cot 0 = root 7,
show that
cosec?e - sec 0 3
cosec?0 + sec?0 4
Answers
Givencotθ=
7
−−(1)
LHS =\frac{(cosec^{2}\theta-sec^{2}\theta)}{(cosec^{2}\theta+sec^{2}\theta)}LHS=
(cosec
2
θ+sec
2
θ)
(cosec
2
θ−sec
2
θ)
=\frac{\frac{1}{sin^{2}\theta}-\frac{1}{cos^{2}\theta}}{\frac{1}{sin^{2}\theta}+\frac{1}{cos^{1}\theta}}=
sin
2
θ
1
+
cos
1
θ
1
sin
2
θ
1
−
cos
2
θ
1
Multiply numerator and denominator by cos^{2}\thetacos
2
θ ,we get
=\frac{\frac{cos^{2}\theta}{sin^{2}\theta}-\frac{cos^{2}\theta}{cos^{2}\theta}}{\frac{cos^{2}\theta}{sin^{2}\theta}+\frac{cos^{2}\theta}{cos^{1}\theta}}=
sin
2
θ
cos
2
θ
+
cos
1
θ
cos
2
θ
sin
2
θ
cos
2
θ
−
cos
2
θ
cos
2
θ
=\frac{cot^{2}\theta-1}{cot^{2}\theta +1}=
cot
2
θ+1
cot
2
θ−1
=\frac{(\sqrt{7})^{2}-1}{(\sqrt{7})^{2}+1}=
(
7
)
2
+1
(
7
)
2
−1
/* From(1)*/
\begin{gathered}=\frac{7-1}{7+1}\\=\frac{6}{8}\\=\frac{3}{4}\\=RHS\end{gathered}
=
7+1
7−1
=
8
6
=
4
3
=RHS
•••♪