Question

75. If cos A = tan B, cos B = tan C and cos C = tan A, then sin A is equal to​

Answer 1

Step-by-step explanation:

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

A2A

[math]\cos A=\tan B\ldots (1)[/math]

[math]\cos B=\tan C \ldots (2)[/math]

[math]\cos C=\tan A \ldots (3)[/math]

We shall find the value of [math]\:\cos B\:[/math]and [math]\:\tan C \:[/math]in terms of [math]\:\sin A\:[/math]from equations (1) and (3) respectively and then substitute those values in equation (2) thus eliminating C and B and creating an equation in [math]\:\sin A[/math]

[math](1)[/math]

[math]\implies \tan B=\cos A[/math]

[math]\implies \dfrac{\sqrt{1-\cos^2 B}}{\cos B}=\cos A[/math]

[math]\implies \dfrac{1-\cos^2 B}{\cos ^2 B}=\cos^2 A=1-\sin^2 A[/math]

[math]\implies\cos^2 B=\dfrac{1}{2-\sin^2 A}\ldots (4)[/math]

[math](3)[/math]

[math]\implies \cos C=\tan A[/math]

Now[math]\:[/math][math]\tan^2 C=\dfrac{1-\cos^2 C}{\cos^2 C}[/math]

[math]\implies \tan^2 C=\dfrac{1-\tan^2 A}{\tan^2 A}[/math]

[math]\implies \tan^2 C=\dfrac{1-2\sin^2 A}{\sin^2 A}\ldots (5)[/math]

Using [math]\:(4)\:[/math]and[math]\:(5)\:[/math]in[math]\:(2)\:[/math]we have-

[math]\dfrac{1}{2-\sin^2 A}=\dfrac{1-2\sin^2 A}{\sin^2 A}[/math]

Let,[math]\:\sin A=x\:[/math]

Thus the equation becomes-

[math]\dfrac{1}{2-x^2}=\dfrac{1-2x^2}{x^2}[/math]

[math]\implies x^4-3x^2+1=0\:[/math]

which is quadratic in[math]\:x^2[/math]

Hence by applying quadratic formula we get,

[math]\implies x^2=\frac{3\pm\sqrt{5}}{2}[/math]

[math]\implies \boxed{\sin A=\pm\sqrt{\frac{3\pm\sqrt{5}}{2}}}[/math]

Some problem this is. Hopefully i got it right. Again my inability to use the math function easily makes me turn towards the good old solving on paper. So here it is....

What is the formula of cosA + cosB?

If tanA=cosB, tanB=cosC, and tanC=cosA, then the value of cosA independent A, B, C?

If cosA=tanB, cosB=tanC, and cosC=tanA, then how do you prove that sinA=sinB=sinC?

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.

Answer 2

Step-by-step explanation:

if helps mark as brainliest