The value of sec2 49-cot2 41- is equal to

Sec^2 49^∘ cot^2 41^∘ =sec^2 49^∘ cot^2( 90^∘ 49^∘) =sec^2 49^∘ tan^249^∘ [∵ cot (90^∘ θ)=tanθ] =1 [∵ 1+tan^2θ=sec^2θ] Hence, the correct answer is option d. ...

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Standard X

Mathematics

Trigonometric Identities

The value of ...

Question

The value of $s e c^{2} 49^{\circ} - c o t^{2} 41^{\circ}$ is equal to

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2 s e c^{2} 49^{\circ}

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2 c o t^{2} 41^{\circ}

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Solution

The correct option is D
1
$s e c^{2} 49^{\circ} - c o t^{2} 41^{\circ} = s e c^{2} 49^{\circ} - c o t^{2} (90^{\circ} - 49^{\circ}) = s e c^{2} 49^{\circ} - t a n^{2} 49^{\circ}$
$[∵ c o t (90^{\circ} - θ) = t a n θ] = 1$
$[∵ 1 + t a n^{2} θ = s e c^{2} θ]$

Hence, the correct answer is option d.

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Similar questions

Q. The value of

sec 41^{o} sin 49^{o} + cos 49^{o} cosec 41^{o}

is :

Q. Value of

{[\frac{sin 49^{o}}{cos 41^{o}} + \frac{cos 41^{o}}{sin 49^{o}}]}^{2}

Q. If

\sin θ = \frac{3}{4}

, prove that

\sqrt{\frac{{cosec}^{2} θ - \cot^{2} θ}{\sec^{2} θ - 1}} = \frac{\sqrt{7}}{3}

Q. The value of

\frac{tan 49^{o}}{cot 41^{o}}

is :

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

(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

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The value of sec2 49∘−cot2 41∘ is equal to

The correct option is D 1sec249∘−cot241∘=sec249∘−cot2(90∘−49∘)=sec2 49∘−tan249∘ [∵ cot (90∘−θ)=tanθ]=1 [∵ 1+tan2θ=sec2θ] Hence, the correct answer is option d.

The value of $s e c^{2} 49^{\circ} - c o t^{2} 41^{\circ}$ is equal to