SCERT Maharashtra solutions for Geometry (Mathematics 2) [English] 10 Standard SSC chapter 6 - Trigonometry [Latest edition]

Choose the correct alternative:

cos θ. sec θ = ?

Choose the correct alternative:

sec 60° = ?

Choose the correct alternative:

1 + cot²θ = ? 

Choose the correct alternative:

cot θ . tan θ = ?

Choose the correct alternative:

sec²θ – tan²θ =?

sin²θ + sin²(90 – θ) = ?

Choose the correct alternative:

`(1 + cot^2"A")/(1 + tan^2"A")` = ?

Choose the correct alternative:

sin θ = `1/2`, then θ = ?

Choose the correct alternative:

tan (90 – θ) = ?

Choose the correct alternative:

cos 45° = ?

Choose the correct alternative:

If sin θ = `3/5`, then cos θ = ?

Choose the correct alternative:

Which is not correct formula?

Choose the correct alternative:

If ∠A = 30°, then tan 2A = ?

`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?

If tan θ = `13/12`, then cot θ = ?

Prove that `"cosec"  θ xx sqrt(1 - cos^2theta)` = 1

If tan θ = 1, then sin θ . cos θ = ?

If 2 sin θ = 3 cos θ, then tan θ = ?

If cot( 90 – A ) = 1, then ∠A = ?

If 1 – cos²θ = `1/4`, then θ = ?

Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`

If tan θ × A = sin θ, then A = ?

(sec θ + tan θ) . (sec θ – tan θ) = ?

`(sin 75^circ)/(cos 15^circ)` = ?

Prove that cos²θ . (1 + tan²θ) = 1. Complete the activity given below.

Activity:

L.H.S = `square`

= `cos^2theta xx square    .....[1 + tan^2theta = square]`

= `(cos theta xx square)^2`

= 1²

= 1

= R.H.S

`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.

Activity:

`5/(sin^2theta) - 5cot^2theta`

= `square (1/(sin^2theta) - cot^2theta)`

= `5(square - cot^2theta)   ......[1/(sin^2theta) = square]`

= 5(1)

= `square`

If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ

Activity:

`square` = 1 + tan²θ    ......[Fundamental trigonometric identity]

`square` – tan²θ = 1

(sec θ + tan θ) . (sec θ – tan θ) = `square`

`sqrt(3)*(sectheta - tan theta)` = 1

(sec θ – tan θ) = `square`

If tan θ = `9/40`, complete the activity to find the value of sec θ.

Activity:

sec²θ = 1 + `square`     ......[Fundamental trigonometric identity]

sec²θ = 1 + `square^2`

sec²θ = 1 + `square` 

sec θ = `square` 

If cos θ = `24/25`, then sin θ = ?

Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ

Prove that `1/("cosec"  theta - cot theta)` = cosec θ + cot θ

If cos(45° + x) = sin 30°, then x = ?

If tan θ + cot θ = 2, then tan²θ + cot²θ = ?

Prove that sec²θ + cosec²θ = sec²θ × cosec²θ

Prove that cot²θ × sec²θ = cot²θ + 1

If 3 sin θ = 4 cos θ, then sec θ = ?

If sin 3A = cos 6A, then ∠A = ?

Prove that sec²θ − cos²θ = tan²θ + sin²θ

Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`

Prove that `(sintheta + tantheta)/cos theta` = tan θ(1 + sec θ)

Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ

Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`

sin⁴A – cos⁴A = 1 – 2cos²A. For proof of this complete the activity given below.

Activity:

L.H.S = `square`

 = (sin²A + cos²A) `(square)`

= `1 (square)`       .....`[sin^2"A" + square = 1]`

= `square` – cos²A    .....[sin²A = 1 – cos²A]

= `square`

= R.H.S

tan²θ – sin²θ = tan²θ × sin²θ. For proof of this complete the activity given below.

Activity:

L.H.S = `square`

= `square (1 - (sin^2theta)/(tan^2theta))`

= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`

= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`

= `tan^2theta (1 - square)`

= `tan^2theta xx square`    .....[1 – cos²θ = sin²θ]

= R.H.S

If tan θ = `7/24`, then to find value of cos θ complete the activity given below.

Activity:

sec²θ = 1 + `square`    ......[Fundamental tri. identity]

sec²θ = 1 + `square^2`

sec²θ = 1 + `square/576`

sec²θ = `square/576`

sec θ = `square` 

cos θ = `square`     .......`[cos theta = 1/sectheta]`

To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.

Activity:

L.H.S = `square`

= `square/sintheta + sintheta/costheta`

= `(cos^2theta + sin^2theta)/square`

= `1/(sintheta*costheta)`     ......`[cos^2theta + sin^2theta = square]`

= `1/sintheta xx 1/square`

= `square`

= R.H.S

If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ

If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ

Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec"  theta)` = sec θ

Prove that cot²θ – tan²θ = cosec²θ – sec²θ 

Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)² 

Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ

Prove that `sec"A"/(tan "A" + cot "A")` = sin A

Prove that `(sintheta + "cosec"  theta)/sin theta` = 2 + cot²θ

Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1

Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A

Prove that sin⁴A – cos⁴A = 1 – 2cos²A

Prove that sec²θ – cos²θ = tan²θ + sin²θ

Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`

In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?

Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`

If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A

Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B

Prove that

sin²A . tan A + cos²A . cot A + 2 sin A . cos A = tan A + cot A

Prove that

sec²A – cosec²A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

Prove that

`(cot "A" + "cosec  A" - 1)/(cot"A" - "cosec  A" + 1) = (1 + cos "A")/"sin A"`

Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ

If cos A = `(2sqrt("m"))/("m" + 1)`, then prove that cosec A = `("m" + 1)/("m" - 1)`

If sec A = `x + 1/(4x)`, then show that sec A + tan A = 2x or `1/(2x)`

In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin²A + sin²B + sin²C = 2, then ∠A = ? , ∠B = ?, ∠C = ?

Prove that sin⁶A + cos⁶A = 1 – 3sin²A . cos²A

Prove that 2(sin⁶A + cos⁶A) – 3(sin⁴A + cos⁴A) + 1 = 0

Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1

If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3

If cos A + cos²A = 1, then sin²A + sin4 A = ?

If cosec A – sin A = p and sec A – cos A = q, then prove that `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1

Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`

If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1

If tan θ – sin²θ = cos²θ, then show that sin² θ = `1/2`.

Prove that (1 – cos²A) . sec²B + tan²B(1 – sin²A) = sin²A + tan²B