Question

In a right-angled triangle, it is given that A is an acute angle and tan A= 5/12.

Find the value of cos+sin

cos−sin​

Answer 1

EXPLANATION.

Right angled triangle, A is an accute angle.

Tan A = 5/12.

→ By using the Pythagorean Theorem,

we get,

→ H² = P² + B²

→ Tan ø = P/B = Perpendicular/Base.

→ Tan A = 5/12 = P/B.

→ H² = (5)² + (12)²

→ H² = 25 + 144

→ H² = 169

→ H² = √169 = 13.

→ Sin A = P/H = Perpendicular/Hypotenuse

Sin A = 5/13.

→ Cos A = B/H = Base/Hypotenuse = 12/13.

→ Tan A = P/B = Perpendicular/Base = 5/12.

→ Csc A = H/P = Hypotenuse/Perpendicular

Csc A = 13/5.

→ Sec A = H/B = Hypotenuse/Base = 13/12.

→ Cot A = B/P = Base/Perpendicular = 12/5.

→ (1) = Cos A + Sin A

→ 12/13 + 5/13 = 17/13.

→ (2) = Cos A - Sin A

→ 12/13 - 5/13 = 7/13

Answer 2

$\huge{\boxed{\bold{if ∅→1st Quadrant}}}$

tan∅=12/5

=Perp/Base

(hyp)²=(b)²+(p)²

=(hyp)²=(5)²+(12)²

=(hyp)²=25+144

=(hyp)²=169

=(hyp)=13

$\huge{\boxed{\bold{sin∅=perp/hyp=12/13}}}$

$\huge{\boxed{\bold{cos∅=base/hyp=5/13}}}$

SO,

sin∅-cos∅=12/13-5/13

=12-5/13

=7/13