Question

if sin theta =k,0<k<1,theta€Q,then tan theta=​

Answer 1

Step-by-step explanation:

sin(θ)=

1

−

1−k

2

,tan(θ)=

k

−

1−k

2

Step-by-step explanation:

1) Since,

k is positive and hence cosine of theta.

=>\thetaθ is in fourth quadrant.

2)

sin(\theta)sin(θ) and tan(\theta)tan(θ) are negative.

We know that,

\begin{gathered}sin^2(\theta)+cos^2(\theta)=1\\ \\= > sin^2(\theta)+k^2=1\\ \\= > sin^2(\theta)=1-k^2\\ \\= > sin(\theta)=-\sqrt{1-k^2}\end{gathered}

sin

2

(θ)+cos

2

(θ)=1

=>sin

2

(θ)+k

2

=1

=>sin

2

(θ)=1−k

2

=>sin(θ)=−

1−k

2

3) Also,

\begin{gathered}1+tan^2(\theta)=sec^2(\theta)\\ \\= > tan^2(\theta)=sec^2(\theta)-1\\ \\= > tan^2(\theta)=\frac{1}{cos^2(\theta)}-1=\frac{1}{k^2}-1=\frac{1-k^2}{k^2}\\ \\ = > tan(\theta)=\frac{-\sqrt{1-k^2}}{k}\end{gathered}

1+tan

2

(θ)=sec

2

(θ)

=>tan

2

(θ)=sec

2

(θ)−1

=>tan

2

(θ)=

cos

2

(θ)

1

−1=

k

2

1

−1=

k

2

1−k

2

=>tan(θ)=

k

−

1−k

2

Hence,

\boxed{tan(\theta)=\frac{-\sqrt{1-k^2}}{k},sin(\theta)=-\sqrt{1-k^2}}

tan(θ)=

k

−

1−k

2

,sin(θ)=−

1−k

2