Which of the following is not a trigonometric identity?

s i n^{2} θ + c o s^{2} θ = 1

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s e c^{2} θ - c o s e c^{2} θ = 1

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c o t^{2} θ + 1 = c o s e c^{2} θ

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s e c^{2} θ - t a n^{2} θ = 1

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Solution

The correct option is B $s e c^{2} θ - c o s e c^{2} θ = 1$
The trigonometric identities are :

$s i n^{2} θ + c o s^{2} θ = 1$

$s e c^{2} θ - t a n^{2} θ = 1$

$c o t^{2} θ + 1 = c o s e c^{2} θ$

Hence, $s e c^{2} θ - c o s e c^{2} θ = 1$ is not a trigonometric identity.

Suggest Corrections

Similar questions

(i) s e c^{2} θ + c o s e c^{2} θ = s e c^{2} θ c o s e c^{2} θ

(i i) t a n^{2} θ - s i n^{2} θ = t a n^{2} θ s i n^{2} θ

(i i i) t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ c o s e c^{2} θ

\sqrt{s e c^{2} θ + c o s e c^{2} θ} = t a n θ + c o t θ

{sec}^{2} θ + c o s e c^{2} θ = {sec}^{2} θ \times c o s e c^{2} θ

(t a n θ + c o t θ)^{2} = s e c^{2} θ + c o s e c^{2} θ

\frac{c o s^{2} θ + t a n^{2} θ - 1}{s i n^{2} θ} = t a n^{2} θ

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