SSC (English Medium) 10th Standard - Maharashtra State Board Important Questions

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

The value of 2tan45° – 2sin30° is ______.

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.

Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`

Find the value of sin 0° + cos 0° + tan 0° + sec 0°.

Find the value of sin 45° + cos 45° + tan 45°.

Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

What will be the value of sin 45° + `1/sqrt(2)`?

In the given figure, if sin θ = `7/13`, which angle will be θ?

Prove that: cot θ + tan θ = cosec θ·sec θ

Proof: L.H.S. = cot θ + tan θ

= `square/square + square/square`  ......`[∵ cot θ = square/square, tan θ = square/square]`

= `(square + square)/(square xx square)`  .....`[∵ square + square = 1]`

= `1/(square xx square)`

= `1/square xx 1/square`

= cosec θ·sec θ  ......`[∵ "cosec"  θ = 1/square, sec θ = 1/square]`

= R.H.S.

∴ L.H.S. = R.H.S.

∴ cot θ + tan θ = cosec·sec θ

If sec θ = `1/2`, what will be the value of cos θ?

Find will be the value of cos 90° + sin 90°.

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`

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`