Question

If cos X=tanY cosY=tanZ cosZ =tanX then sinx=siny=sinz

Answer 1

Question :- if cos x = tan y cos y= tan z cos z = tan x then sin x = sin y = sin z = ?

Solution :-

→ cos x = tan y

Squaring both sides,

→ cos² x = tan² y

using tan²A = (sec²A - 1) in RHS now,

→ cos² x = (sec² y - 1) ----------- Eqn(1)

Also,

→ cos y = tan z

using cosA = (1/secA) and tanA = (1/cotA) ,

→ (1/sec y) = 1/(cot z)

→ sec y = cot z --------------- Eqn(2)

Putting value of Eqn(2) in Eqn(1) Now,

→ cos² x = (cot² z - 1)

→ 1 + cos² x = cot² z

using cotA = (cosA /sinA) Now,

→ 1 + cos² x = (cos² z / sin² z)

Now, using sin²A = (1 - cos²A) , in RHS denominator,

→ 1 + cos² x = cos² z / ( 1 - cos² z )

Now, we have given that, cos z = tan x

So,

→ 1 + cos² x = tan² x / ( 1 - tan² x )

using cos²A = (1 - sin²A) in LHS, and tanA = (sinA/cosA) in RHS , we get,

→ 1 + ( 1 - sin² x ) = ( sin² x / cos² x ) / {1 - (sin² x / cos² x)}

→ 2 - sin² x = sin² x / ( cos² x - sin² x )

using cos²A - sin²A = 1 - 2sin²A in RHS now,

→ 2 - sin² x = sin² x / ( 1 - 2 sin² x )

Cross - Multiply,

→ ( 2 - sin² x )( 1 - 2 sin² x ) = sin² x

→ 2 - 4sin² x - sin²x - sin²x + 2sin⁴ x = 0

→ 2 sin⁴ x - 6 sin² x + 2 = 0

→ 2(sin⁴ x - 3 sin² x + 1) = 0

→ sin⁴ x - 3 sin² x + 1 = 0

Now, Let, sin² x = M

Than,

→ M² - 3M + 1 = 0

using, Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

→ x = [ -b±√(b²-4ac) / 2a ]

Here,

• a = 1
• b = (-3)
• c = 1

→ M = [ - (-3) ± √{(-3)² - 4*1*1} ] / 2*1

→ M = [ 3 ± √(9-4) ] / 2

→ M = ( 3 ± √5 ) / 2

Therefore,

→ sin² x = (3 ± √5) / 2

But, ( 3 + √5 ) /2 > 1 and sin² x ≤ 1.

Hence,

→ sin² x = ( 3 - √5 ) / 2

→ sin² x = ( 6 - 2√5 ) / 4

→ sin² x = ( 5 + 1 - 2 * √5 * 1 ) / 4

→ sin² x = {(√5)² + (1)² - 2 * √5 * 1} / 4

comparing RHS numerator with a² + b² - 2ab = (a - b)² ,

→ sin² x = (√5 - 1)² / 4

Square - root both sides now,

→ sin x = ( √5 - 1 ) / 2

Multiply and divide by 2 both numerator and denominator of RHS now,

→ sin x = 2( √5 - 1 ) / (2 * 2)

→ sin x = 2( √5 - 1 ) / 4

→ sin x = 2[ ( √5 - 1 ) / 4 ]

using value of sin 18° in RHS now,

→ sin x = 2 sin 18°

Hence, similarly we can show that sin x = sin y = sin z = 2 sin 18° = ( √5 - 1 ) / 2 ...