Question

If 3 tan A = 4 then prove that
sec A-cosec A 1
(i)
sec A+ cosec A 7
1-sin A 1
1+cos A 2√2
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Answer 1

$\underline{\underline{\maltese\: \: \textbf{\textsf{Proper question:-}}}}$

If 3tan A = 4 then prove that :

(i) √secA-cosecA/√secA+cosecA = 1/√3

(ii) 1−sinA/ 1+cos A = 1 /2√2

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

3tanA=4

or, tanA=4/3=p/b

Using Pythagorus theorem, h²-p²+²b we get,

⟹h²=4²2+3²

⟹h²=16+9

⟹h²=25

or, h=5 (neglecting the negative sign)

secA=h/b=5/3 and cosecA=h/p=5/4

I) secA-cosecA/secA+cosecA

⟹√(5/3-5/4)/(5/3+5/4)

⟹√(5/12)/(35/12)

⟹ √(5/12x12/35)

⟹√1/7

ii) 1-sinA/1+cosA

⟹ (1-4/5)/(1+3/5)

⟹{(5-4)/5}{(5+3)/5)

⟹(1/5)/(8/5)

⟹1/5x5/8

Answer 2

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