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or,cot theta=1/tan theta=1/3/4=4/3
or,cot^2theta=16/9
Now, sin theta=(1/cosec^2 theta)^1/2
=[1/(1+cot^2 theta)]^1/2
So,sin theta=[1/(1+16/9)]^1/2
=[1/(16+9)/9]^1/2
=(9/25)^1/2
or,sin theta=3/5
Again, sec theta=1/cos theta
or,sec theta=[1/(1-sin^2 theta)]^1/2
or,sec theta=[1/(1-9/25)]^1/2
or, sec theta=[25/(25-9)]^1/2
or, sec theta=(25/16)^1/2
or, sec theta=5/4
Ans: sin theta=3/5 & sec theta=5/4
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Ved Prakash Sharma, former Lecturer at Sbm Inter College, Rishikesh (1971-2007)
Answered 2 years ago · Author has 9.3K answers and 6.1M answer views
Let theta = x
tan x=3/4. , sin x=? and. sec x=?
sec^2x=1+tan^2x=1+9/16= 25/16
sec x=+/-(5/4). Answer
cos x= +/- 4/5
sin x=+/-√(1-cos^2x) = +/- √(1–16/25)=+/- √(9/25) = +/-3/5. Answer.
Second-method:-
tan x= 3/4
or. sin x/cos x= 3/4
or. (sin x)/3= (cos x)/4 =k(led)
or. sin x= 3k. , cos x=4k
sin^2x+cos^2x=1. =>. 9k^2+16k^2=1. or. 25k^2 = 1
k = +/-1/5
sin x= 3k = +/-3/5. Answer
and. cos x=4k = +/-(4/5). or sec x= +/-(5/4). Answer.