Question

Qno. 1 ) Write sin theta in terms of cos theta.

Q 2 ) Write sec theta in terms of cot theta.

Q 3 ) Write tan theta in terms of sec theta.


Q 4 ) Write Cot theta in terms of cos theta.

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Answer 1

Q 1

By trigonometric identity we know that $sin^2\theta+cos^2\theta=1$ .

$sin^2\theta+cos^2\theta=1$

Transpose $cos^2\theta$ to the other side :

$\implies sin^2\theta=1-cos^2\theta$

Take square root both sides and we get :

$\implies \boxed{sin\theta=\sqrt{1-cos^2\theta}}$

Q 2

We also know that $sec^2\theta-tan^2\theta=1$

$sec^2\theta-tan^2\theta=1$

We know that $tan\theta-\frac{1}{cot\theta}$

$\implies sec^2\theta-\frac{1}{cot^2\theta}=1$

$\implies sec^2\theta=1+\frac{1}{cot^2\theta}$

$\implies sec^2\theta=\frac{cot^2\theta+1}{cot^2\theta}$

Take square root both sides :

$\implies \boxed{sec\theta=\frac{\sqrt{cot^2\theta+1}}{cot\theta}}$

Q 3

By trigonometric identities we have :

$sec^2\theta-tan^2\theta=1$

Transpose the value of $tan^2\theta$ to the other side :

$\implies tan^2\theta=sec^2\theta-1$

Take square roots on both sides :

$\implies \boxed{tan\theta=\sqrt{sec^2\theta-1}}$

Q 4 :

We know that $cosec^2\theta-cot^2\theta=1$

$\implies cot^2\theta=cosec^2\theta-1$

Use $cosec^2\theta=\frac{1}{sin^2\theta}$

$\implies cot^2\theta=\frac{1}{sin^2\theta}-1$

$\implies cot^2\theta=\frac{1-sin^2\theta}{sin^2\theta}$

Use $1-sin^2\theta=cos^2\theta$ and $sin^2\theta=1-cos^2\theta$

$\implies cot^2\theta=\frac{cos^2\theta}{1-cos^2\theta}$

Take square root both sides :-

$\implies \boxed{cot\theta=\frac{cos\theta}{\sqrt{1-cos^2\theta}}}$

Answer 2

Trigonometric formulas :

      • sinx * cosecx = 1

      • cosx * secx = 1

      • tanx * cotx = 1

      • sin²x + cos²x = 1

      • sec²x - tan²x = 1

      • cosec²x - cot²x = 1

Solution :

1)

We know that,

         sin²θ + cos²θ = 1

    or, sin²θ = 1 - cos²θ

    or, sinθ = $\sqrt{1-cos^{2}\theta}$ ,

which is the required term presentation.

2)

We know that,

         sec²θ - tan²θ = 1

    or, sec²θ = 1 + tan²θ

    or, sec²θ = 1 + $\frac{1}{cot^{2}\theta}$ , since tanx*cotx = 1

    or, secθ = $\sqrt{(1+\frac{1}{cot^{2}\theta})}$

    or, secθ = $\sqrt{\frac{cot^{2}\theta+1}{cot^{2}\theta}}$

which is the required term presentation.

3)

We know that,

         sec²θ - tan²θ = 1

    or, tan²θ = sec²θ - 1

    or, tanθ = $\sqrt{sec^{2}\theta-1}$ ,

which is the required term presentation.

4)

We know that,

         cosec²θ - cot²θ = 1

    or, cot²θ = cosec²θ - 1

    or, cot²θ = $\frac{1}{sin^{2}\theta}-1$

    or, cot²θ = $\frac{1}{1-cos^{2}\theta}-1$

    or, cot²θ = $\frac{1-1+cos^{2}\theta}{1-cos^{2}\theta}$

    or, cotθ = $\sqrt{\frac{1-1+cos^{2}\theta}{1-cos^{2}\theta}}$

    or, cotθ = $\sqrt{\frac{cos^{2}\theta}{1-cos^{2}\theta}}$

which is the required term presentation.