Prove sinx/cos3x + sin3x/cos9x +sin9x/cos27x
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Step-by-step explanation:
ANSWER
cos3x
sinx
+
cos9x
sin3x
+
cos27x
sin9x
multiplying and dividing by 2
=
2
1
[
cos3x
2sinx
+
cos9x
2sin3x
+
cos27x
2sin9x
]
multiplying and dividing by cosx,cos9x,cos27 respectively and simplifying
=
2
1
[
cos3xcosx
sin2x
+
cos9xcos3x
sin6x
+
cos27xcos9x
sin18x
]
Use, sin(A−B)=sinAcosB−cosAsinB
=
2
1
[
cos3xcosx
sin3xcosx−cos3xsinx
]+
2
1
[
cos27xcos9x
sin9xcos3x−cos9xsin3x
]
+
2
1
[
cos27xcos9x
sin27xcos9x−cos27xsin9x
]
=
2
1
[tan27x−tan9x+tan9x−tan3x+tan3x−tanx]=
2
1
[tan27x−tanx]
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