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Trigonometric Ratios of Standard Angles

Question 2 ii...

Question

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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Solution

(ii) Option (D) is correct.
$\frac{(1 - t a n^{2} 45^{\circ})}{(1 + t a n^{2} 45^{\circ})} = \frac{(1 - 1^{2})}{(1 + 1^{2})} = \frac{0}{2} = 0$

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Similar questions

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}} =

t a n 90^{\circ}

s i n 45^{\circ}

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

\frac{2 t a n 3 0^{0}}{1 + t a n^{2} 30^{0}} =

s i n 60^{\circ}

c o s 60^{\circ}

t a n 60^{\circ}

s i n 30^{\circ}

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

\frac{2 t a n 3 0^{0}}{1 - t a n^{2} 30^{0}} =

c o s 60^{\circ}

s i n 60^{\circ}

t a n 60^{\circ}

s i n 30^{\circ}

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°

\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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Trigonometric Ratios of Standard Angles

Standard IX Mathematics

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