the solution of equation sec theta -cosec theta =4/3
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sec@-cosec@=4/3,
1/cos@ - 1/sin@=4/3,
(sin@-cos@)/[email protected]@ =4/3,
(sin@-cos@)=4sin@cos@/3,
S.B.S
(sin@-cos@)²=16sin²@.cos²@/9,
sin²@+cos²@[email protected]@=16sin²@.cos²@/9,
1-2sin@cos@=16sin²@.cos²@/9,
9-18sin@cos@=16([email protected]@)²,
16([email protected]@)²[email protected]@-9=0,
16([email protected]@)² [email protected]@[email protected]@-9=0,
[email protected]@([email protected]@+3)-3([email protected]@+3)=0,
([email protected]@+3)([email protected]@-3)=0,
take
[email protected]@-3=0
[email protected]@=3/4,
sin2@=3/4,
then
2@=[sin^(-1) 3/4],
@=1/2 [sin^(-1) 3/4]
sec@-cosec@=4/3,
1/cos@ - 1/sin@=4/3,
(sin@-cos@)/[email protected]@ =4/3,
(sin@-cos@)=4sin@cos@/3,
S.B.S
(sin@-cos@)²=16sin²@.cos²@/9,
sin²@+cos²@[email protected]@=16sin²@.cos²@/9,
1-2sin@cos@=16sin²@.cos²@/9,
9-18sin@cos@=16([email protected]@)²,
16([email protected]@)²[email protected]@-9=0,
16([email protected]@)² [email protected]@[email protected]@-9=0,
[email protected]@([email protected]@+3)-3([email protected]@+3)=0,
([email protected]@+3)([email protected]@-3)=0,
take
[email protected]@-3=0
[email protected]@=3/4,
sin2@=3/4,
then
2@=[sin^(-1) 3/4],
@=1/2 [sin^(-1) 3/4]
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Answered by
5
Answer:
Step-by-step explanation:sec@-cosec@=4/3,
1/cos@ - 1/sin@=4/3,
(sin@-cos@)/[email protected]@ =4/3,
(sin@-cos@)=4sin@cos@/3,
S.B.S
(sin@-cos@)²=16sin²@.cos²@/9,
sin²@+cos²@[email protected]@=16sin²@.cos²@/9,
1-2sin@cos@=16sin²@.cos²@/9,
9-18sin@cos@=16([email protected]@)²,
16([email protected]@)²[email protected]@-9=0,
16([email protected]@)² [email protected]@[email protected]@-9=0,
[email protected]@([email protected]@+3)-3([email protected]@+3)=0,
([email protected]@+3)([email protected]@-3)=0,
take
[email protected]@-3=0
[email protected]@=3/4,
sin2@=3/4,
then
2@=[sin^(-1) 3/4],
@=1/2 [sin^(-1) 3/4]
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