Complete the following activity to prove: cotθ + tanθ = cosecθ × secθ Activity: L.H.S. = cotθ + tanθ = `cosθ/sinθ + square/cosθ` = `(square + sin^2theta)/(sinθ xx cosθ)` = `1/(sinθ xx cosθ)` ....... ∵ `square` = `1/sinθ xx 1/cosθ` = `square xx secθ` &there4; L.H.S. = R.H.S.

L.H.S. = cotθ + tanθ = `cosθ/sinθ + sinθ/cosθ` = `(cos^2θ + sin^2theta)/(sinθ xx cosθ)` = `1/(sinθ xx cosθ)` ....... ∵ sin2θ cos2θ = 1 = `1/sinθ xx 1/cosθ` = cosecθ × secθ &there4; L.H.S. = R.H.S.

Complete the following activity to prove: cotθ + tanθ = cosecθ × secθ Activity: L.H.S. = cotθ + tanθ = θθθcosθsinθ+□cosθ = θθ□+sin2θsinθ×cosθ = θθ1sinθ× cosθ ....... ∵ □ = θθ1sinθ×1cosθ = θ□×secθ - Geometry Mathematics 2

Question

Complete the following activity to prove:

cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= $\frac{\cos θ}{\sin θ} + \frac{□}{\cos θ}$

= $\frac{□ + \sin^{2} θ}{\sin θ \times \cos θ}$

= $\frac{1}{\sin θ \times \cos θ}$ ....... ∵ $□$

= $\frac{1}{\sin θ} \times \frac{1}{\cos θ}$

= $□ \times \sec θ$

∴ L.H.S. = R.H.S.

Fill in the Blanks

Sum

Solution

L.H.S. = cotθ + tanθ

= $\frac{\cos θ}{\sin θ} + \frac{\sin θ}{\cos θ}$

= $\frac{\cos^{2} θ + \sin^{2} θ}{\sin θ \times \cos θ}$

= $\frac{1}{\sin θ \times \cos θ}$ ....... ∵ sin²θ cos²θ = 1

= $\frac{1}{\sin θ} \times \frac{1}{\cos θ}$

= cosecθ × secθ

∴ L.H.S. = R.H.S.

shaalaa.com

Trigonometric Identities

Is there an error in this question or solution?

Q 2.A.iii Q 2.A.ii Q 2.B.i

2021-2022 (March) Set 1

APPEARS IN

2021-2022 (March) Set 1 (with solutions)

Q 2.A.iii | 2 marks

Video TutorialsVIEW ALL [5]

RELATED QUESTIONS

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \cos e c A + \cot A$ using the identity cosec² A = 1 cot²A.

Show that $\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \cos e c A + \cot A$

Prove the following trigonometric identities.

$\frac{\tan A}{{(1 + \tan^{2} A)}^{2}} + \frac{\cot A}{(1 + \cot^{2} A)} = \sin A \cos A$

Prove the following trigonometric identities.

$\frac{\cot^{2} A (\sec A - 1)}{1 + \sin A} = \sec^{2} A (\frac{1 - \sin A}{1 + \sec A})$

Prove the following trigonometric identities.

$\frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B$

Prove the following identities:

$\frac{\sec A - 1}{\sec A + 1} = \frac{1 - \cos A}{1 + \cos A}$

Prove the following identities:

$\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^{2} A - 1}$

$\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 \cos e c θ$

Write the value of $(1 + \cot^{2} θ) \sin^{2} θ$ .

Prove the following identity :

secA(1 - sinA)(secA + tanA) = 1

Prove the following identity :

$\sqrt{\cos e c^{2} q - 1} = cosq cosecq$

If sec θ = x + $\frac{1}{4 x}, x \neq 0,$ find (sec θ + tan θ)

Prove that $\frac{\cos θ}{1 - \sin θ} = \frac{1 + \sin θ}{\cos θ}$ .

Prove that: 2(sin⁶θ + cos⁶θ) - 3 ( sin⁴θ + cos⁴θ) + 1 = 0.

Prove that $\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{\cos A}$ .

Prove that $\frac{\tan^{3} θ}{1 + \tan^{2} θ} + \frac{\cot^{3} θ}{1 + \cot^{2} θ} = \sec θ . \cos e c θ - 2 \sin θ \cos θ .$

If $\frac{\cos θ}{1 + \sin θ} = \frac{1}{a}$ , then prove that $\frac{a^{2} - 1}{a^{2} + 1}$ = sin θ

Given that sin θ = $\frac{a}{b}$ , then cos θ is equal to ______.

If cos (α + β) = 0, then sin (α – β) can be reduced to ______.

(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec²θ.

Complete the following activity to prove: cotθ + tanθ = cosecθ × secθ Activity: L.H.S. = cotθ + tanθ = θθθcosθsinθ+□cosθ = θθ□+sin2θsinθ×cosθ = θθ1sinθ× cosθ ....... ∵ □ = θθ1sinθ×1cosθ = θ□×secθ - Geometry Mathematics 2

Advertisements

Advertisements

Question

Solution

APPEARS IN

Video TutorialsVIEW ALL [5]

RELATED QUESTIONS

Select a course

Complete the following activity to prove: cotθ + tanθ = cosecθ × secθ Activity: L.H.S. = cotθ + tanθ = θθθcosθsinθ+□cosθ = θθ□+sin2θsinθ×cosθ = θθ1sinθ× cosθ ....... ∵ □ = θθ1sinθ×1cosθ = θ□×secθ - Geometry Mathematics 2

Advertisements

Advertisements

Question

SolutionShow Solution

APPEARS IN

Video TutorialsVIEW ALL [5]

RELATED QUESTIONS

Select a course

Solution