Express sinA cosA tanA secA cotA in terms of cosecA
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We have : ( Sin A + Cos A ) ( tan A + Cot A ) = Sec A + Cosec A
Taking LHS :
( Sin A + Cos A ) ( tan A + Cot A )
⇒Sin A tan A + Sin A Cot A + Cos A tan A + Cos A Cot A⇒Sin A ×Sin ACos A + Sin A ×Cos ASin A + Cos A ×Sin ACos A + Cos A ×Cos ASin A ( we know tan θ = Sin θCos θ and Cot θ = Cos θSin θ )⇒Sin 2ACos A +Cos A + Sin A +Cos2 ASin A⇒Sin 2ACos A +Cos A + Sin A +Cos2 ASin A⇒Sin 2A + Cos2 ACos A + Sin 2A + Cos2 ASin A⇒1Cos A + 1Sin A ( we know Sin 2θ + Cos2 θ = 1 )⇒Sec A +Cosec A ( we know Sec θ = 1Cos θ and Cosec θ = 1Sin θ )
Therefore,
L.H.S. = R.H.S. ( Hence proved )
⭐️⭐️⭐️⭐️hope this helps you
We have : ( Sin A + Cos A ) ( tan A + Cot A ) = Sec A + Cosec A
Taking LHS :
( Sin A + Cos A ) ( tan A + Cot A )
⇒Sin A tan A + Sin A Cot A + Cos A tan A + Cos A Cot A⇒Sin A ×Sin ACos A + Sin A ×Cos ASin A + Cos A ×Sin ACos A + Cos A ×Cos ASin A ( we know tan θ = Sin θCos θ and Cot θ = Cos θSin θ )⇒Sin 2ACos A +Cos A + Sin A +Cos2 ASin A⇒Sin 2ACos A +Cos A + Sin A +Cos2 ASin A⇒Sin 2A + Cos2 ACos A + Sin 2A + Cos2 ASin A⇒1Cos A + 1Sin A ( we know Sin 2θ + Cos2 θ = 1 )⇒Sec A +Cosec A ( we know Sec θ = 1Cos θ and Cosec θ = 1Sin θ )
Therefore,
L.H.S. = R.H.S. ( Hence proved )
⭐️⭐️⭐️⭐️hope this helps you
ayushi2503:
but my question is different and your answer is different
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