Question

if cos A 9/2 then find value of (1) cos C tan A +sin C cos A (2) cos C tanA +tan C cos A​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

If CosA=tanB, cosB=tanC cosC =tanA then the numerical value of sin A =?

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A2A

cosA=tanB…(1)

cosB=tanC…(2)

cosC=tanA…(3)

We shall find the value of cosB and tanC in terms of sinA from equations (1) and (3) respectively and then substitute those values in equation (2) thus eliminating C and B and creating an equation in sinA

(1)

⟹tanB=cosA

⟹1−cos2B−−−−−−−−√cosB=cosA

⟹1−cos2Bcos2B=cos2A=1−sin2A

⟹cos2B=12−sin2A…(4)

(3)

⟹cosC=tanA

Now tan2C=1−cos2Ccos2C

⟹tan2C=1−tan2Atan2A

⟹

cos C=tan A

so tan C=sqrt(1-tan ^2 A)/tan A

cos B=tan C=sqrt(1-tan ^2 A)/tan A

so tan B=sqrt((2×tan ^2 A-1)/(1-tan ^2 A))

But cos A=tan B

cos A=sqrt(2×tan ^2 A-1)/(1-tan ^2 A))

cos ^2 A=(2×tan ^2 A-1)/(1-tan ^2 A)

1/sec ^2 A=(2×tan ^2 A-1)/(1-tan ^2 A)

1/(1+tan ^2 A)=(2×tan ^2 A-1)/(1-tan ^2 A)

tan ^4 A +tan ^2 A -1=0

tan ^2 A=(-1+√5)/2

cot ^2 A=2/(√5 -1)

cosec ^2 A -1 =2/(√5 -1)

cosec ^2 A=(1+√5)/(√5 -1)

sin ^2 A=(√5–1)/(√5+1)=(√5–1)^2/(√5+1)(√5–1)

sin ^2 A=((√5–1)^2)/4

sin A=(√5–1)/2.