SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

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

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

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)`.

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 + sin⁴A = ?

If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(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 `sin^2θ = 1/2`.

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

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`

∴ L.H.S. = R.H.S.

If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.