prove that: cosec 2A+ cot 4A= cot A - cosec 4A
Answers
Answer:
The answer is simple
Step-by-step explanation:
Identities Used:
\begin{gathered}\sf{cosecA=\dfrac{1}{sinA}}\\\sf{sin2A=2sinAcosA}\\\sf{\dfrac{cosA}{sinA}=cotA}\\\sf{cos2A=2cos^2A-1}\end{gathered}
cosecA=
sinA
1
sin2A=2sinAcosA
sinA
cosA
=cotA
cos2A=2cos
2
A−1
Solution:
\sf{Taking\:L.H.S.}TakingL.H.S.
\sf{=cosec2A+cosec4A}=cosec2A+cosec4A
\sf{=\dfrac{1}{sin2A}+\dfrac{1}{sin4A}}=
sin2A
1
+
sin4A
1
\sf{=\dfrac{1}{sin2A}+\dfrac{1}{2sin2Acos2A}}=
sin2A
1
+
2sin2Acos2A
1
\sf{Taking\:L.C.M.}TakingL.C.M.
\sf{=\dfrac{2cos2A+1}{2sin2Acos2A}}=
2sin2Acos2A
2cos2A+1
\sf{=\dfrac{2cos2A+1+cos4A-cos4A}{sin4A}}=
sin4A
2cos2A+1+cos4A−cos4A
\sf{=\dfrac{2cos2A+1+cos4A}{sin4A}-\dfrac{cos4A}{sin4A}}=
sin4A
2cos2A+1+cos4A
−
sin4A
cos4A
\sf{=\dfrac{2cos2A+1+2cos^2A-1}{sin4A}-cot4A}=
sin4A
2cos2A+1+2cos
2
A−1
−cot4A
\sf{=\dfrac{2cos2A(1+cos2A)}{2sin2Acos2A}-cot4A}=
2sin2Acos2A
2cos2A(1+cos2A)
−cot4A
\sf{=\dfrac{1+2cos^2A-1}{sin2A}-cot4A}=
sin2A
1+2cos
2
A−1
−cot4A
\sf{=\dfrac{2cos^2A}{2sinAcosA}-cot4A}=
2sinAcosA
2cos
2
A
−cot4A
\sf{=\dfrac{cosA}{sinA}-cot4A}=
sinA
cosA
−cot4A
\sf{=cotA-cot4A}=cotA−cot4A
\bf{\sf{HENCE\:PROVED}}HENCEPROVED