Math, asked by dularranu, 5 months ago

Simplify: (cosec A – sin A)(sec A – cos A)(tan A + cot A)

dularranu: cosA is answer or not

Answers

Answered by tennetiraj86

Answer:

answer for the given problem is 1

Using formulae:-

Cosec A=1/Sin A
Sec A=1/Cos A
Tan A=Sin A/Cos A
Cot A=Cos A/Sin A
Sin² A+Cos² A=1

Attachments:

dularranu: f a cos A + b sin A= 4 and a sin A− b cos A= 3, then evaluate a^2 + b^2
.

tennetiraj86: Given equations are :acosA+bSinA=4 on squaring both sides then

tennetiraj86: (acosA+bSinA)²=4² =>a²Cos2A+2abSinAcosA+b²Sin²A----(1)

tennetiraj86: and a SinA-bCos A=3 ,on squaring both sides then

tennetiraj86: (aSinA-bCosA)²=3², a²Sin²A-2absinAcosA+b²Cos²A---(2)

tennetiraj86: adding(1)&(2) then,we get,a²Sin²A+a²Cos²A+b²sin²A+b²Cos²A=16+9,

tennetiraj86: a²(sin²A+Cos²A)+b²(Sin²A+cos²A)=25

tennetiraj86: a²(1)+b²(1)=25

tennetiraj86: a²+b²=25

dularranu: thanks bro

Answered by sandy1816

$(coseca - sina)(seca - cosa)(tana + cota) \\ \\ = ( \frac{1}{sina} - sina)( \frac{1}{cosa} - cosa)( \frac{sina}{cosa} + \frac{cosa}{sina} ) \\ \\ = ( \frac{1 - {sin}^{2}a }{sina} )( \frac{1 - {cos}^{2}a }{cosa} )( \frac{ {sin}^{2}a + {cos}^{2} a}{sinacosa} ) \\ \\ = \frac{ {sin}^{2} a {cos}^{2}a }{ {sin}^{2} a {cos}^{2} a} \\ \\ = 1$

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Simplify: (cosec A – sin A)(sec A – cos A)(tan A + cot A)