If tan A = 5/12, find the value of (sin A + cos A) sec A.
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Hii friend,
Tan A = 5/12 => P/B
P = 5 , B = 12
By pythagoras theroem,
(H)² = (B)² + (P)²
(H)² = (12)² + (5)²
(H)² = 144+25
H = ✓169 = 13
Therefore,
Sin A = P/H => 5/13
And,
Cos A = B/H => 12/13
Sec A = H/B => 13/12
(SinA + CosA ) SecA
(5/13 + 12/13) 13/12
=> (5+12/13) × 13/12
=> 17/13 × 13/12
=> 17/12
HOPE IT WILL HELP YOU...... :-)
Tan A = 5/12 => P/B
P = 5 , B = 12
By pythagoras theroem,
(H)² = (B)² + (P)²
(H)² = (12)² + (5)²
(H)² = 144+25
H = ✓169 = 13
Therefore,
Sin A = P/H => 5/13
And,
Cos A = B/H => 12/13
Sec A = H/B => 13/12
(SinA + CosA ) SecA
(5/13 + 12/13) 13/12
=> (5+12/13) × 13/12
=> 17/13 × 13/12
=> 17/12
HOPE IT WILL HELP YOU...... :-)
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