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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

1 -tan2 45∘ 1...

Question

$\frac{1 - \tan^{2} 45 °}{1 + \tan^{2} 45 °}$ is equal to

(a) tan 90°
(b) 1
(c) sin 45°
(d) sin 0°

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Solution

We have to find the value of the following

So

We know that

Hence the correct option is

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Similar questions

Choose the correct option and justify your choice.

(i)

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii)

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv)

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

Q. Question 2 (ii)
Choose the correct option and justify your choice.
(ii)

\frac{1 - t a n^{2} 45^{0}}{1 + t a n^{2} 45^{0}} =

(A)

t a n 90^{\circ}

(B) 1
(C)

s i n 45^{\circ}

(D) 0

\frac{1 + \tan^{2} A}{1 + \cot^{2} A}

is equal to

(a) sec² A
(b) −1
(c) cot² A
(d) tan² A

Choose the correct alternative answer for following multiple choice questions.

(i) Which of the following statements is true ?

(A) sin

θ

= cos (90

-

θ

) (B) cos

θ

= tan (90

-

θ

)
(C) sin

θ

= tan (90

-

θ

) (D) tan

θ

= tan (90

-

θ

)

(ii) Which of the following is the value of sin 90^°?

(A)

\frac{\sqrt{3}}{2}

(B) 0

(C)

\frac{1}{2}

(D) 1

(iii) 2 tan 45^°+ cos 45^°

-

sin 45^°= ?

(A) 0 (B) 1

(iv)

\frac{\cos 28^{°}}{\sin 62^{°}}

(A) 2 (B)

-

1 (C) 0 (D) 1

Q. Question 2 (ii)
Choose the correct option and justify your choice.
(ii)

\frac{1 - t a n^{2} 45^{0}}{1 + t a n^{2} 45^{0}} =

(A)

t a n 90^{\circ}

(B) 1
(C)

s i n 45^{\circ}

(D) 0

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1-tan2 45°1+tan2 45° is equal to (a) tan 90° (b) 1 (c) sin 45° (d) sin 0°

We have to find the value of the following So We know that Hence the correct option is

$\frac{1 - \tan^{2} 45 °}{1 + \tan^{2} 45 °}$ is equal to

(a) tan 90°
(b) 1
(c) sin 45°
(d) sin 0°

We have to find the value of the following

So

We know that

Hence the correct option is