1/ ( 1+ sin thrita ) + 1/ (1-sin thiita) = 2 sec^2 thrita
Answers
Answered by
0
Step-by-step explanation:
LHS =left hand side, RHS =right hand side
LHS
=
1
1
+
sin
θ
+
1
1
−
sin
θ
=
1
−
sin
θ
+
1
+
sin
θ
(
1
+
sin
θ
)
(
1
−
sin
θ
)
->Common Denominator
=
1
−
sin
θ
+
1
+
sin
θ
(
1
+
sin
θ
)
(
1
−
sin
θ
)
=
2
1
−
sin
2
x
=
2
cos
2
x
=
2
⋅
1
cos
2
x
=
2
sec
2
x
=
R
H
S
Answered by
0
Answer:
# 1/(1+ sin theta ) + 1 / (1-sin theta )
by using identify
#. (a-b) (a+b) = a square - b square
so we apply identify in denominator
so we
# 1 /(1 square - sin theta square )
# 1/ 1- sin square theta
as we know
# 1- sin square theta = cos square theta
so now
# 1/ cos square theta
so we 1/ cos theta = sec theta
# 1 / cos square theta = sec square theta
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