Question

Given sinA=3/4 find cosA,tanA,cotA and secA​

Answer 1

EXPLANATION.

sin A = 3/4. = p/h = perpendicular/hypotenuse.

By using Pythagorean theorem we get,

= H² = B² + P²

= (4)² = B² + (3)²

= 16 = B² + 9

= B² = 7

= B = √7

Sin A = p/h = 3/4.

Cos A = b/h = √7/4

Tan A = p/b = 3/√7

cosec A = h/p = 4/3

sec A = h/b = 4/√7

Cot A = b/p = √7/3

More information.

= Sin²A + Cos²A = 1

= Tan²A + 1 = Sec²A

= 1 + Cot²A = Csc²A

= Sin(2A) = 2sinAcosA

= Cos(2A) = Cos²A - Sin²A

= Sin²(A/2) = 1 - CosA/2

= Cos²(A/2) = 1 + CosA/2

Answer 2

$\bigstar \sf \bf \underline{\underline{Given:-}}$

• Sin A = 3/4.

$\bigstar \sf \bf \underline{\underline{To\ find:-}}$

• Cos A ,tan A,cot A & sec A.

$\bigstar \sf \bf \underline{\underline{Solution:-}}$

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Let us take a ΔABC , right angled at C.

We know :

❥Sinθ=Opposite side/Hypotenuse

Acc to question ,

=> Sin A = 3/4.

Let Opposite side = 3x & hypotenuse = 4x.

To find the adjacent side :

From pythagoras theorem ,

❥ AB² = BC²+AC²

⇒ (4x)² = (9x)² + AC²

⇒ 16x² = 9x²+AC²

⇒ 16x²- 9x² =AC²

⇒ 7x² = AC²

⇒ AC = √7 x.

Therefore, Adjacent side = √7 x.

Now,

→ Cos A = Adj / Hypotenuse

               = √7 x / 4x

               = √7 / 4.

→ Tan A = Opp / Adj

              = 3x /√7 x

               = 3/√7.

→ Cot A = Adj / opp

              = √7 x/3x

             = √7 / 3

→ Sec A = Hypotenuse / Adj

              = 4x  / √7 x.

               = 4/√7.

-----------------------------------

$\bigstar \underbrace{\sf \bf More\ shots:-}$

✦ Cot A = 1/tan A

✦ Sec A = 1/cos A

✦ Sin²A+Cos²A = 1

✦ Tan A = sin A / cos A

✦ Cot A =  Cos A / sin A

✦ Sec²A - tan²A = 1

✦ Cosec²A - cot²A = 1

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HOPE IT HELPS !