Question

if cos a= tan b, cos b = tanc and cos c = tan a
prove that sin a= sin b= sin c

Answer 1

Answer:

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

Ad by Sundaram Mutual

Are you a conservative investor?

Invest in Balanced Advantage Fund. Dynamically allocates across asset classes based on market conditions.

Start Now

5 Answers



Zafar Hussain, ISB & NIT alumnus

Answered November 25, 2015

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....





Continue Reading

Related Questions (More Answers Below)

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

655 Views

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

1,031 Views

In a triangle if cosA=cosB*cosC then what is value of tanA-tanB-tanC?

4,851 Views

What is the formula of sinA + cosB?

36,598 Views

What is the formula of cosA + cosB?

7,004 Views

Other Answers



Rhea Saikia, Lady Shri Ram College for Women

Answered June 17, 2017 · Author has 148 answers and 246.2k answer views

A2A

cosA=tanB…(1)cos⁡A=tan⁡B…(1)

cosB=tanC…(2)cos⁡B=tan⁡C…(2)

cosC=tanA…(3)cos⁡C=tan⁡A…(3)

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 [/math]

Continue Reading

Sponsored by State Bank of India

Avail customized packages.

Opt for the corporate salary package for your employees and enjoy a range of privileges.

Learn More



Vivek Singh, studying computer science & engineering in nit jamshedpur

Updated April 22, 2017 · Author has 253 answers and 390.2k answer views

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.