If sec A + tan A is equal to 3 and sec A - tan A is equal to 1 by 3 find the value of sin A and cot A + cosec A
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Given that... secA+tanA=3 and secA-tanA=1/3
From here, we can write that
(secA+tanA)/(secA-tanA)=9
Appkying componendo dividendo we get
secA/tanA=10/8
So....... ¦¦¦¦¦¦¦¦¦¦¦¦[sinA=4/5] ¦¦¦¦¦¦¦¦¦¦¦¦
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********CotA+cosecA=(1+cosA) /sinA
*******cosA=under root(1-sin²A)=3/5
*******So required answer is (1+(3/5))/(4/5)
::::::::::The final value is 2::::::::::
From here, we can write that
(secA+tanA)/(secA-tanA)=9
Appkying componendo dividendo we get
secA/tanA=10/8
So....... ¦¦¦¦¦¦¦¦¦¦¦¦[sinA=4/5] ¦¦¦¦¦¦¦¦¦¦¦¦
----------------------------------------------------------------------
********CotA+cosecA=(1+cosA) /sinA
*******cosA=under root(1-sin²A)=3/5
*******So required answer is (1+(3/5))/(4/5)
::::::::::The final value is 2::::::::::
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