sin(tan⁻¹(tan7π/6))+cos(cos⁻¹(cos7π/3)=.......,Select Proper option from the given options.
(a) -1
(b) 0
(c) 1
(d) √3/2
Answers
Answered by
5
we have to find the value of sin[tan^-1(tan7π/6)]+cos[cos^-1(cos7π/3)]
we know, tan^-1(tanx) = x for , -π/2 < x < π/2
so, tan^-1(tan7π/6) = tan^-1[tan(π + π/6)]
= tan^-1[tan(π/6)] [ as we know, tan(π + x) = tanx]
= tan^-1(tanπ/6) = π/6
hence, tan^-1(tan7π/6) = π/6 .....(1)
again, cos^-1(cosx) = x , for -π/2 ≤ x ≤ π/2
so, cos^-1(cos7π/3) = cos^-1[cos(2π+π/3)]
= cos^-1[cos(π/3)] [as we know, cos(2π + x) = cosx]
= π/3
hence, cos^-1(cos7π/6) = π/3 ........(2)
now, from equations (1) and (2),
= sin[π/6 + π/3]
= sinπ/2
= 1
therefore option (c) is correct
we know, tan^-1(tanx) = x for , -π/2 < x < π/2
so, tan^-1(tan7π/6) = tan^-1[tan(π + π/6)]
= tan^-1[tan(π/6)] [ as we know, tan(π + x) = tanx]
= tan^-1(tanπ/6) = π/6
hence, tan^-1(tan7π/6) = π/6 .....(1)
again, cos^-1(cosx) = x , for -π/2 ≤ x ≤ π/2
so, cos^-1(cos7π/3) = cos^-1[cos(2π+π/3)]
= cos^-1[cos(π/3)] [as we know, cos(2π + x) = cosx]
= π/3
hence, cos^-1(cos7π/6) = π/3 ........(2)
now, from equations (1) and (2),
= sin[π/6 + π/3]
= sinπ/2
= 1
therefore option (c) is correct
Answered by
4
HELLO DEAR,
sin[tan-¹(tan7π/6)] + cos[cos-¹(cos7π/3)]
we know:- tan-¹ (tanx) = x
so,
sin(7π/6) + cos(7π/3)
AND sin(π + x) = sinx and cos(2π + x) = cosx
therefore, sin(π + π/6) + cos(2π + π/3)
sinπ/6 + cosπ/3
(1/2) + (1/2)
(1 + 1)/2
2/2
1
hence, option (c) is correct.
I HOPE ITS HELP YOU DEAR,
THANKS
sin[tan-¹(tan7π/6)] + cos[cos-¹(cos7π/3)]
we know:- tan-¹ (tanx) = x
so,
sin(7π/6) + cos(7π/3)
AND sin(π + x) = sinx and cos(2π + x) = cosx
therefore, sin(π + π/6) + cos(2π + π/3)
sinπ/6 + cosπ/3
(1/2) + (1/2)
(1 + 1)/2
2/2
1
hence, option (c) is correct.
I HOPE ITS HELP YOU DEAR,
THANKS
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