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# If θ is an acute angle and sin θ = cos θ, find the value of 2 tan^2 θ + sin^2 θ – 1 - CBSE Class 10 - Mathematics

ConceptTrigonometric Ratios of Some Specific Angles

#### Question

If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1

#### Solution

sin θ = cos θ

\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{\cos \theta }{\cos\theta }

[Dividing both sides by cos θ]

⇒ tanθ = 1

⇒ tanθ = tan45° ⇒ θ= 45°

∴ 2 tan^2 θ + sin^2 θ – 1

= 2tan^2 45° + sin^2 45° – 1

=2(2)^{2}+( \frac{1}{\sqrt{2}} )^{2}-1

=2+\frac{1}{2}-1=\frac{5}{2}-1=\frac{3}{2}

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Solution If θ is an acute angle and sin θ = cos θ, find the value of 2 tan^2 θ + sin^2 θ – 1 Concept: Trigonometric Ratios of Some Specific Angles.
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