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SSC (Marathi Semi-English) इयत्ता १० वी - Maharashtra State Board Question Bank Solutions

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

Prove that sec2θ + cosec2θ = sec2θ × cosec2θ

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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Prove that cot2θ × sec2θ = cot2θ + 1

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

If 3 sin θ = 4 cos θ, then sec θ = ?

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

Prove that sec2θ − cos2θ = tan2θ + sin2θ

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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]`

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

Prove that cot2θ – tan2θ = cosec2θ – sec2θ 

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined

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

[6] Trigonometry
Chapter: [6] Trigonometry
Concept: undefined >> undefined
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