prove that secant theta + tan theta minus secant theta minus tan theta is equals to 1 + 2 tan squared theta + 2 Sin Theta into tan theta
akashpriya27:
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tan
2
θ
+
1
=
sec
2
θ
Therefore,
L
H
S
=
(
(
sec
θ
+
tan
θ
)
2
−
1
)
(
(
sec
θ
+
tan
θ
)
2
+
1
)
=
sec
2
θ
+
tan
2
θ
+
2
sec
θ
tan
θ
−
1
sec
2
θ
+
tan
2
θ
+
2
sec
θ
tan
θ
+
1
=
2
tan
2
θ
+
2
sec
θ
tan
θ
2
sec
2
θ
+
2
sec
θ
tan
θ
=
2
tan
θ
tan
θ
+
sec
θ
2
sec
θ
tan
θ
+
sec
θ
=
tan
θ
sec
θ
=
sin
θ
cos
θ
⋅
cos
θ
=
sin
θ
=
R
H
S
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