SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions

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 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 `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)`

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