In the given figure, find 3 tan θ – 2 sin α + 4 cos α.

First of all, we find the value of RS

In right angled ∆RQS, we have

(RQ)² + (QS)² = (RS)²

⇒ (8)² + (6)² = (RS)²

⇒ 64 + 36 = (RS)²

⇒ RS =√100

⇒ RS =±10 [taking positive square root, since side cannot be negative]

⇒ RS =10

Now, we find the value of QP

In right angled ∆RQP

(RQ)² + (QP)² = (RP)²

⇒ (8)² + (QP)² = (17)²

⇒ 64 + (QP)² = 289

⇒ (QP)² =289–64

⇒ (QP)² =225

⇒ QP =√225

⇒ QP =±15 [taking positive square root, since side cannot be negative]

⇒ QP =15

tan θ

Now, 3 tan θ – 2 sinα + 4cos α

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