prove that please it is urgent
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hey there is answer !!!
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now start your answer !!!!
_______________________________
Dear Student,
Please find below the correct solution to the asked query:
Given : tan θ1 − cotθ + cot θ1 − tan θ = 1 + Sec θ Cosec θ
Taking L.H.S.
⇒ tan θ1 − cotθ + cot θ1 − tan θ
⇒Sin θcos θ1 − cos θSin θ+cos θSin θ1 −Sin θcos θ ( As we know tan θ = Sin θCos θ And Cot θ = Cos θSin θ )
⇒Sin θcos θSin θ − cos θSin θ + cos θSin θcos θ − Sin θcos θ
⇒Sin θ×Sin θcos θSin θ − cos θ + cos θ×cos θSin θcos θ − Sin θ
⇒Sin2 θcos θSin θ − cos θ + cos2 θSin θcos θ − Sin θ
⇒Sin2 θcos θSin θ − cos θ + cos2 θSin θ−(Sin θ − cos θ)
⇒Sin2 θcos θSin θ − cos θ − cos2 θSin θSin θ − cos θ
⇒Sin2 θcos θ − cos2 θSin θSin θ − cos θ
⇒Sin3 θ − cos3 θSin θ cos θ (Sin θ − cos θ)
⇒ (Sin θ − cos θ)(Sin2 θ + cos2 θ +Sin θ cos θ )Sin θ cos θ (Sin θ − cos θ) [As we know ( a3 - b3 ) = ( a - b ) ( a2 + b 2 + ab ) ]
⇒(Sin2 θ + cos2 θ +Sin θ cos θ )Sin θ cos θ
⇒ (1 +Sin θ cos θ )Sin θ cos θ ( As we know sin2 θ + cos2 θ = 1 )
⇒ 1 Sin θ cos θ + Sin θ cos θSin θ cos θ
⇒ 1 + Sec θ Cosec θ ( As we know Secθ = 1Cos θ And Cosec θ = 1Sin θ )
Hence
L.H.S. = R.H.S. ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
by.......
smart abhishek ..........
========================
I hope you help !!!
===========================
mark be brainlist frd ......
=============================
now start your answer !!!!
_______________________________
Dear Student,
Please find below the correct solution to the asked query:
Given : tan θ1 − cotθ + cot θ1 − tan θ = 1 + Sec θ Cosec θ
Taking L.H.S.
⇒ tan θ1 − cotθ + cot θ1 − tan θ
⇒Sin θcos θ1 − cos θSin θ+cos θSin θ1 −Sin θcos θ ( As we know tan θ = Sin θCos θ And Cot θ = Cos θSin θ )
⇒Sin θcos θSin θ − cos θSin θ + cos θSin θcos θ − Sin θcos θ
⇒Sin θ×Sin θcos θSin θ − cos θ + cos θ×cos θSin θcos θ − Sin θ
⇒Sin2 θcos θSin θ − cos θ + cos2 θSin θcos θ − Sin θ
⇒Sin2 θcos θSin θ − cos θ + cos2 θSin θ−(Sin θ − cos θ)
⇒Sin2 θcos θSin θ − cos θ − cos2 θSin θSin θ − cos θ
⇒Sin2 θcos θ − cos2 θSin θSin θ − cos θ
⇒Sin3 θ − cos3 θSin θ cos θ (Sin θ − cos θ)
⇒ (Sin θ − cos θ)(Sin2 θ + cos2 θ +Sin θ cos θ )Sin θ cos θ (Sin θ − cos θ) [As we know ( a3 - b3 ) = ( a - b ) ( a2 + b 2 + ab ) ]
⇒(Sin2 θ + cos2 θ +Sin θ cos θ )Sin θ cos θ
⇒ (1 +Sin θ cos θ )Sin θ cos θ ( As we know sin2 θ + cos2 θ = 1 )
⇒ 1 Sin θ cos θ + Sin θ cos θSin θ cos θ
⇒ 1 + Sec θ Cosec θ ( As we know Secθ = 1Cos θ And Cosec θ = 1Sin θ )
Hence
L.H.S. = R.H.S. ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
by.......
smart abhishek ..........
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