Q10. Find the value of a and b from the following:
3
5-18
1
3+ V8
= a - bv8
Answers
Step-by-step explanation:
CORRECT QUESTION :-
\begin{gathered}\\ \red \bigstar \displaystyle \tt \: prove \: that \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta \\ \\\end{gathered}
★provethat
1+sinθ
cosθ
+
1−sinθ
cosθ
=2secθ
GIVEN :-
\begin{gathered}\\ \red \bigstar \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta \\ \\\end{gathered}
★
1+sinθ
cosθ
+
1−sinθ
cosθ
=2secθ
TO PROVE :-
\begin{gathered}\\ \red \bigstar \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta \\ \\\end{gathered}
★
1+sinθ
cosθ
+
1−sinθ
cosθ
=2secθ
SOLUTION :-
\begin{gathered}\\ \red \bigstar \: \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta \\\end{gathered}
★
1+sinθ
cosθ
+
1−sinθ
cosθ
=2secθ
Take L.H.S Part,
\begin{gathered}\\ \implies \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } \\ \\\end{gathered}
⟹
1+sinθ
cosθ
+
1−sinθ
cosθ
\begin{gathered}\blue \bigstar \orange{ \tt \: Taking \: L.C.M \bigg(1 + \sin \theta\bigg)\bigg(1 - \sin \theta\bigg)} \\ \\\end{gathered}
★TakingL.C.M(1+sinθ)(1−sinθ)
\begin{gathered}\implies \displaystyle \tt \: \dfrac{ \cos \theta\bigg(1 - \sin \theta \bigg) + \cos \theta \bigg(1 + \sin \theta \bigg)} { \bigg(1 + \sin \theta \bigg) \bigg(1 - \sin \theta \bigg) } \\ \\\end{gathered}
⟹
(1+sinθ)(1−sinθ)
cosθ(1−sinθ)+cosθ(1+sinθ)
\begin{gathered}\displaystyle \tt \implies \: \dfrac{ \cos \theta - \cos \theta. \sin \theta + \cos \theta + \cos \theta. \sin \theta}{\bigg(1 + \sin \theta\bigg)\bigg(1 - \sin \theta\bigg)} \\ \\\end{gathered}
⟹
(1+sinθ)(1−sinθ)
cosθ−cosθ.sinθ+cosθ+cosθ.sinθ
\begin{gathered}\blue \bigstar \orange{\tt Using \: identity :- (a + b)(a -b) = a^{2} - b^{2}} \\ \\\end{gathered}
★Usingidentity:−(a+b)(a−b)=a
2
−b
2
\begin{gathered}\implies \displaystyle \tt \: \dfrac{ \cos \theta + \cos \theta}{(1) ^{2} - \sin ^{2} \theta } \\ \\\end{gathered}
⟹
(1)
2
−sin
2
θ
cosθ+cosθ
\begin{gathered}\implies \displaystyle \tt \: \dfrac{2 \cos \theta}{1 - \sin ^{2} \theta } \\ \\\end{gathered}
⟹
1−sin
2
θ
2cosθ
\begin{gathered}\blue \bigstar \orange{\tt Using \: identity :- 1 - \sin ^{2} \theta = \cos ^{2} \theta } \\ \\\end{gathered}
★Usingidentity:−1−sin
2
θ=cos
2
θ
\begin{gathered}\implies \displaystyle \tt \: \dfrac{2\cos \theta}{ \cos ^{2}\theta} \\ \\\end{gathered}
⟹
cos
2
θ
2cosθ
\begin{gathered}\implies \displaystyle \boxed{\tt \:2 \sec \theta} \\ \\\end{gathered}
⟹
2secθ
L.H.S = R.H.S
HENCE VERIFIED ✅
Answer:
nice question 5-8
NX. jkcnf