find tan θ if cos θ = 1/6 and sin θ<0...?

1. -sqrt 35

2. 1

3. -1

4.

(1 - sinx)/cosx = cos x / (1+Sinx)

(1+sinx)(1-sinx)=cosx.cosx

(1-sin²x)=cos²X and cos²x=1-sin²x then

1-sin²x=1-sin²x

1. Cos x is positive and sine is negative, so the angle is in quadrant III. Draw a reference triangle with x = 1 and hypotenuse 6 and use pythagorean theorem to find

that y = sqrt35. Tanx = y/x and is negative in QIII so tan x = -sqrt35.

2. cosecant is the reciprocal of the sine, and the product of any number and its reciprocal is 1. (The -x is a smokescreen, but don't forget that sine is an odd function).

3. cos(pi/2 -x) = sin x (cofunctions) and csc(-x) = -1/sinx (reciprocal relation and odd function) so the product is -1.

4. You have a lot of options here; one is to multiply the numerator and denominator on the left side by (1 + sin t). This gives you

(1 - sin^2t)/[cost(1 + sin t)] but 1 - sin^t2 is cos^2t (pythagorean identity)

so you have cos^2t/[cost(1 + sin t)] = cos t/( 1 + sin t) (one of the cost cancels)

1. sin θ is < 0 in Quadrants 3 and 4, becomes sin represents the y axis. If cos is 1/6, we know its positive, and the only quadrant cosine is positive in is 1 and 4. so, we are in quadrant 4.

cos1/6 = adj/hyp

c^2=a^2+b^2 = 36=1^2+b^2

36=1+b^2

35=b^2

=+-sqrt35 = b.

tangent is opp/adj; therefore it is B, tangent is negative in the 4th quadrant.

2. csc(-x)sin(-x) can be rewritten as 1/sin(-x) sin(-x) = when multiplied sin(-x)/sin(-x) = b. 1.

3. cos(pi/2-x) is the same as the sin(x). sin(x)csc(-x) = 1/csc(x)csc(-x) = -1.

not sure on the last 1.

1. cos is adjacent/hyp., so adjacent side must be 1, and hyp. must be 6

Therefore, (using pythagorus in a right angled triangle) the oppostie side must be sqrt. of (6^2 - 1^2) = sqrt. 35

for sinθ to be less than 0 the answer must be negative ( you can see this by tping it onto your calculator)

=-sqrt35

2. csc(-x)sin(-x) = 1/sin(-x) (sin(-x)) = sin(-x)/sin(-x) = 1

3. cos(pi/2-x)csc(-x) = cos(pi/2)cos(-x) - sin(pi/2)sin(-x) (1/sin(-x))

cos(pi/2) = 0, sin(pi/2) = 1:

0(csc(-x)) -1(sin(-x)) (1/sin(-x))

=(sin(-x))(1/sin(-x))

=sin(-x)/sin(-x)

=1

4.Sorry not sure about this one :/

1.(b)

2.(b)

3.(a)

4.(1-cos t)(1+cos t)/(cos t)1+cos t

=1-sin^2/(cos t)1+cos t

=cos^2 t/cos t(1+cos t)

because 1-sin^2= cos^2 now cancelling cos t from up & down.

=(cos t / 1 + sin t)

sin²θ + cos²θ = 1

sin²θ + (1/6)² = 1

sin²θ = 1 - 1/36

sin²θ = 35/36

sinθ = ±√(35)/6 (negative root because sinθ < 0)

tanθ = sinθ / cosθ

tanθ = -√(35) / 1

tanθ = -√(35)

-----------------

csc(-x)sin(-x)

1/sin(-x) * sin(-x)

1

----------------------

cos(π/2 - x) * csc(-x)

sin(x) * 1/sin(-x)

sin(x) * -1/sin(x)

-1

--------------------

(1 - sinθ) / cosθ = cosθ / (1 + sinθ)

(1 - sinθ) / cosθ = cosθ(1 - sinθ) / [(1 + sinθ)(1 - sinθ)]

(1 - sinθ) / cosθ = cosθ(1 - sinθ) / (1 - sin²θ)

(1 - sinθ) / cosθ = cosθ(1 - sinθ) / cos²θ

(1 - sinθ) / cosθ = (1 - sinθ) / cosθ

1-b) - sqrt 35

2-b) 1

3-a) -1

4-

(1-sin t)(1+sint)

= ------------------

cost(1+sint)

(1-sin^2t)

=-------------

cost(1+sint)

cos^2t

=--------

cost(1+sint)

cost/(1+sint)