Question

B.With the help of a figure prove that. Sin’O+ cos²O =1, where is an acute angle.​

Answer 1

Step-by-step explanation:

Sin²A + Cos²A = 1

to find : -

prove that.

\huge\sf\underline{Solution}

Solution

Consider a right angle triangle.

Note :- Picture is in the attachment.

From that,

SinA = \sf\frac{BC}{AC}

AC

BC

squaring on both sides

Sin²A = \sf\frac{(BC)²}{(AC)²}

(AC)²

(BC)²

----(1)

CosA = \sf\frac{AB}{AC}

AC

AB

squaring on both sides

Cos²A = \sf\frac{(AB)²}{(AC)²}

(AC)²

(AB)²

--- (2)

According to right angle triangle :-

(AC)² = (AB)²+(BC)² ---- (3)

adding equation 1 and 2 :-

Sin²A + Cos²A = \sf\frac{(BC)²}{(AC)²}

(AC)²

(BC)²

+ \sf\frac{(AB)²}{(AC)²}

(AC)²

(AB)²

Sin²A + Cos²A = \sf\frac{(BC)²+(AB)²}{(AC)²}

(AC)²

(BC)²+(AB)²

Sin²A + Cos²A = \sf\frac{(AC)²}{(AC)²}

(AC)²

(AC)²

since, from equation 3.

Sin²A + Cos²A = 1