Tan 2- 2-2- 2-cosec2-

The correct option is A True nbsp;We know that, 1 + tan^2θ = sec^2θ .....(1) and 1 + cot^2θ = cosec^2θ .....(2) Adding (1) and (2), 2 + nbsp;tan^2θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

tan 2θ+ 2θ+2=...

Question

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

True

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False

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Solution

The correct option is A True
We know that, $1 + t a n^{2} θ = s e c^{2} θ . . . . . (1) and 1 + c o t^{2} θ = c o s e c^{2} θ . . . . . (2) Adding (1) a n d (2), 2 + t a n^{2} θ + c o t^{2} θ = s e c^{2} θ + c o s e c^{2} θ$
Hence, the given statement is true.

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Similar questions

Q. Prove that:

(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

(ii)

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

(iii)

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

Q. Solve :

\frac{{cot}^{2} θ + {sec}^{2} θ}{{tan}^{2} θ + {cosec}^{2} θ} = (sin θ cos θ) (tan θ + cot θ)

Q. The minimum value of

{sin}^{2} θ + {cos}^{2} θ + {sec}^{2} θ + c o s e c^{2} θ + {tan}^{2} θ + {cot}^{2} θ

Q. Prove the following identity

\frac{{tan}^{2} θ}{{tan}^{2} θ - 1} + \frac{c o s e c^{2} θ}{{sec}^{2} θ - c o s e c^{2} θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ}

Q. Show that

{tan}^{2} θ + {cot}^{2} θ + 2 = {sec}^{2} θ . {csc}^{2} θ

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tan2θ+cot2θ+2=sec2θ+cosec2θ

The correct option is A True We know that, 1+tan2θ=sec2θ.....(1) and1+cot2θ=cosec2θ.....(2)Adding (1) and (2),2+tan2θ+cot2θ=sec2θ+cosec2θ Hence, the given statement is true.

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

The correct option is A True
We know that, $1 + t a n^{2} θ = s e c^{2} θ . . . . . (1) and 1 + c o t^{2} θ = c o s e c^{2} θ . . . . . (2) Adding (1) a n d (2), 2 + t a n^{2} θ + c o t^{2} θ = s e c^{2} θ + c o s e c^{2} θ$
Hence, the given statement is true.