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Question: If \(4\sin\theta = 3\cos\theta\) then \(\frac{\sec^{2}\theta}{4\lbrack 1 - \tan^{2}\theta\rbrack}\) ...

If 4sin⁡θ=3cos⁡θ4\sin\theta = 3\cos\theta then sec⁡2θ4[1−tan⁡2θ]\frac{\sec^{2}\theta}{4\lbrack 1 - \tan^{2}\theta\rbrack} equals to

A

25/1625/16

B

2528\frac{25}{28}

C

14\frac{1}{4}

D

1

Answer

2528\frac{25}{28}

Explanation

Solution

Given 4sin⁡θ=3cos⁡θ4\sin\theta = 3\cos\theta ⇒ tan⁡θ=34\tan\theta = \frac{3}{4}

The given expression is sec⁡2θ4[1−tan⁡2θ]=1+tan⁡2θ4(1−tan⁡2θ)\frac{\sec^{2}\theta}{4\lbrack 1 - \tan^{2}\theta\rbrack} = \frac{1 + \tan^{2}\theta}{4(1 - \tan^{2}\theta)}

=1+9164(1−916)=2528\frac{1 + \frac{9}{16}}{4\left( 1 - \frac{9}{16} \right)} = \frac{25}{28}.