beginarray - c-c- Whine Column A - Column B Whlinea -sin90- - A - i -cos- A ll hlineb tan90- -circ- A - ii - cosec - A Whlinec 90- circ - A - iii - A lilhlinelend array A- a-i- b-i i- c-i i iB - a - i - c - ii - b - iii C- b - i- c - ii- a - iiiD- c - i- a - ii- b - iii

The correct option is B a (i), c (ii), b (iii) sin(90^∘ – A ) = cosA tan(90^∘ – A) = cotA sec(90^∘ – A) = cosecA ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

beginarray | ...

Question

$\begin{matrix} C o l u m n A & C o l u m n B (a) s i n (90^{\circ} - A) & (i) c o s A (b) t a n (90^{\circ} - A) & (i i) c o s e c A (c) s e c (90^{\circ} - A) & (i i i) c o t A \end{matrix}$

a - (i), b - (ii), c - (iii)

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a - (i), c - (ii), b - (iii)

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b - (i), c - (ii), a - (iii)

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c - (i), a - (ii), b - (iii)

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Solution

The correct option is B
a - (i), c - (ii), b - (iii)

sin( $90^{\circ}$ – A ) = cosA

tan( $90^{\circ}$ – A) = cotA

sec( $90^{\circ}$ – A) = cosecA

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Column AColumn B(a) sin (90∘−A)(i) cos A(b) tan(90∘−A) (ii) cosec A(c) sec(90∘−A)(iii) cot A

The correct option is B a - (i), c - (ii), b - (iii) sin(90∘ – A ) = cosA tan(90∘ – A) = cotA sec(90∘ – A) = cosecA

$\begin{matrix} C o l u m n A & C o l u m n B (a) s i n (90^{\circ} - A) & (i) c o s A (b) t a n (90^{\circ} - A) & (i i) c o s e c A (c) s e c (90^{\circ} - A) & (i i i) c o t A \end{matrix}$

The correct option is B
a - (i), c - (ii), b - (iii)

sin( $90^{\circ}$ – A ) = cosA

tan( $90^{\circ}$ – A) = cotA

sec( $90^{\circ}$ – A) = cosecA