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Plz solve these questions which I have attached the image......

Plz solve these questions which I have attached the image......

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4 Answers

Snehal Jadhav
54 Points
5 years ago
In the given attachment the answer for 1st will be:
    sin(a+b)=1
So,a+b=90  (in degrees).......(i)
    sin(a-b)=1/2
So,a-b=60............(ii)
By solving the above two equations we get
a=60
b=30
Thus, tan(a+2b)=tan(60+2(30))
         =tan(120) =-(3)½
Similarly, tan(2a+b)=tan(2(60)+30)
         =tan(150)=1÷(-(3)1/2 )
tan(a+2b)*tan(2a+b)=1
        Hence, the answer is (A)1.
 
   
Snehal Jadhav
54 Points
5 years ago
Answer for the second question is:
           5cosA+3=0
       cosA=-3/5    &     sinA=-4/5  (BY PYTHAGORAS THEOREM)
\therefore tanA=4/3
The quadratic euation formed will be:
(x-4/5)(x+4/3)=0
(5x-4)(3x+4)=0
15x2 +20x-12x-16=0
15x2 +8x-16=0
 Thus, the answer is (B)15x2+8x-16=0
Laxman Machha
8 Points
5 years ago
Answer for 3rd question is:-
SecA=5x+(1/20x)
=>SecA=(100x2+1)/20x-----------(1)
=>TanA=(100x2-1)/20x------------(2)
(1)+(2)
=>(100x2+1+100x2-1)/20x
=>200x2/20x
=>10x(or)1/10x
So, the correct option is (D)
na batau
17 Points
5 years ago
 
In the given attachment the answer for 1st will be:
    sin(a+b)=1
So,a+b=90  (in degrees).......(i)
    sin(a-b)=1/2
So,a-b=60............(ii)
By solving the above two equations we get
a=60
b=30
Thus, tan(a+2b)=tan(60+2(30))
         =tan(120) =-(3)½
Similarly, tan(2a+b)=tan(2(60)+30)
         =tan(150)=1÷(-(3)1/2 )
tan(a+2b)*tan(2a+b)=1
        Hence, the answer is (A)1.
 
Answer for the second question is:
           5cosA+3=0
       cosA=-3/5    &     sinA=-4/5  (BY PYTHAGORAS THEOREM)
\therefore tanA=4/3
The quadratic euation formed will be:
(x-4/5)(x+4/3)=0
(5x-4)(3x+4)=0
15x2 +20x-12x-16=0
15x2 +8x-16=0
 Thus, the answer is (B)15x2+8x-16=0
 
 
Answer for 3rd question is:-
SecA=5x+(1/20x)
=>SecA=(100x2+1)/20x-----------(1)
=>TanA=(100x2-1)/20x------------(2)
(1)+(2)
=>(100x2+1+100x2-1)/20x
=>200x2/20x
=>10x(or)1/10x
So, the correct option is (D)
 

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