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

If cot θ = `40/9`, find the values of cosec θ and sinθ,

We have, 1 + cot²θ = cosec²θ

1 + `square` = cosec²θ

1 + `square` = cosec²θ

`(square + square)/square` = cosec²θ

`square/square` = cosec²θ  ......[Taking root on the both side]

cosec θ = `41/9`

and sin θ = `1/("cosec"  θ)`

sin θ = `1/square`

∴ sin θ =  `9/41`

The value is cosec θ = `41/9`, and sin θ = `9/41`

In an equilateral triangle PQR, prove that PS² = 3(QS)².

Show that, cotθ + tanθ = cosecθ × secθ

Solution :

L.H.S. = cotθ + tanθ

= `cosθ/sinθ + sinθ/cosθ`

= `(square + square)/(sinθ xx cosθ)`

= `1/(sinθ xx cosθ)` ............... `square`

= `1/sinθ xx 1/square`

= cosecθ × secθ

L.H.S. = R.H.S

∴ cotθ + tanθ = cosecθ × secθ

Eliminate θ if x = r cosθ and y = r sinθ.

`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.

Find the value of sin²θ  + cos²θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB² + BC² = `square`   .....(Pythagoras theorem)

Divide both sides by AC²

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 

In the following figure, m(arc PMQ) = 130^o, find ∠PQS.

In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle if m(arc PCR) = 26°, m(arc QDS) = 48°, then find:
(i) m∠PQR
(ii) m∠SPQ
(iii) m∠RAQ

Write the equation of the line passing through A(–3, 4) and B(4, 5) in the form of ax + by + c = 0

If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tan​θ.

If \[\tan \theta = \frac{3}{4}\], find the values of sec​θ and cos​θ

If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.

If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`

Prove that:

cos²θ (1 + tan²θ)

Prove that:

Prove that:

(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.

Prove that:

Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`