1-tan A- A1- A-cosec A-a 1b 0c 2d 4

(c) 2 (1+tanA+secA)(1+cotA #8722;cosecA)=(1+sinAcosA+1cosA)(1+cosAsinA #8722;1sinA)=(cosA+sinA+1cosA)(sinA+cosA #8722;1sinA)=(cosA+sinA)+1(sinA+cosA) # ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

1+tan A+ A1+ ...

Question

(1 + tan A + sec A)(1 + cot A − cosec A) = ?
(a) 1
(b) 0
(c) 2
(d) 4

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Solution

(c) 2

$(1 + \tan A + s ec A) (1 + c ot A - c o \sec A) = (1 + \frac{sin A}{cos A} + \frac{1}{cos A}) (1 + \frac{cos A}{sin A} - \frac{1}{sin A}) = (\frac{cos A + sin A + 1}{cos A}) (\frac{sin A + c os A - 1}{sin A}) = \frac{{(cos A + sin A) + 1} {(sin A + cos A) - 1}}{cos A sin A} = \frac{{(cos A + s in A)}^{2} - 1}{cos A sin A} = \frac{({cos}^{2} A + {sin}^{2} A + 2 cos A sin A) - 1}{cos A sin A} = \frac{1 + 2 cos A sin A - 1}{cos A sin A} (∵ {cos}^{2} A + {sin}^{2} A = 1) = \frac{2 cos A sin A}{cos A sin A} = 2$

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Similar questions

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(1 + tan A + sec A)(1 + cot A − cosec A) = ? (a) 1 (b) 0 (c) 2 (d) 4

(c) 2 (1+tanA+secA)(1+cotA−cosecA)=(1+sinAcosA+1cosA)(1+cosAsinA−1sinA)=(cosA+sinA+1cosA)(sinA+cosA−1sinA)={(cosA+sinA)+1}{(sinA+cosA)−1}cosAsinA=(cosA+sinA)2−1cosAsinA=(cos2A+sin2A+2cosAsinA)−1cosAsinA=1+2cosAsinA−1cosAsinA (∵cos2A+sin2A=1)=2cosAsinAcosAsinA=2

(1 + tan A + sec A)(1 + cot A − cosec A) = ?
(a) 1
(b) 0
(c) 2
(d) 4