Question

Class - 10

Trignometry

Explain all trignometric ratios in terms of tan θ.​

Answer 1

Things seem easier when we take baby steps. Starting with cot∅ ;-)

→ cot∅ = 1/tan∅

Coming onto sin∅ and cos∅,

→ tan∅ = sin∅/cos∅

→ tan∅ × cos∅ = sin∅

Remember sec∅, I know you forgot. Well, cos∅ = 1/sec∅.

→ sin∅ = tan∅/sec∅

Hoping you remember sec²∅ - tan²∅ = 1, hence sec∅ = √(1 + tan²∅)

→ sin∅ = tan∅/√(1 + tan²∅)

Wait! Before going cos∅, I won't let you forget cosec∅.

→ cosec∅ = 1/sin∅

→ cosec∅ = √(1 + tan²∅)/tan∅

Let's now start with cos∅

→ cos∅ = sin∅/tan∅

We know value of sin∅, ryt?

→ cos∅ = tan∅/√(1 + tan²∅)/tan∅

→ cos∅ = 1/√(1 + tan²∅)

Yes, I won't let you forget sec∅

→ sec∅ = 1/cos∅

→ sec∅ = √(1 + tan²∅)

Hence answered all :-)

Answer 2

Answer:

$\displaystyle{\sin\theta=\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}}\\\\\\\displaystyle{\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}}\\\\\\\displaystyle{\cot\theta=\frac{1}{\tan\theta}}$

$\displaystyle{\sec\theta=\sqrt{1+\tan^2\theta}}$

$\displaystyle{cosec \ \theta=\dfrac{1}{\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}}}$

Step-by-step explanation:

We have  trigonometric identity :

sin² θ + cos² θ = 1

sec² θ - tan² θ = 1

Starting from sec θ :

We have :

sec² θ - tan² θ = 1

sec² θ = 1 + tan² θ

sec θ = √ (  1 + tan² θ )

We know cos θ = 1 / sec θ

$\displaystyle{\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}}$

Now for sin θ :

We have :

$\displaystyle{\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}}$

Square on both side :

$\displaystyle{\cos^2\theta=\dfrac{1}{1+\tan^2\theta}}$

From identity we have :

cos² θ = 1 - sin² θ

$\displaystyle{1-\sin^2\theta=\dfrac{1}{1+\tan^2\theta}}\\\\\\\displaystyle{\sin^2\theta=\dfrac{1+\tan^2\theta-1}{1+\tan^2\theta}}\\\\\\\displaystyle{\sin^2\theta=\dfrac{\tan^2\theta}{1+\tan^2\theta}}\\\\\\\displaystyle{\sin\theta=\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}}$

Now for cosec θ we know :

cosec θ = 1 / sin θ

$\displaystyle{cosec \ \theta=\dfrac{1}{\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}}}$

Now for cot θ we know :

$\displaystyle{\cot\theta=\frac{1}{\tan\theta}}$

Hence we get all answer.