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`        Plz solve these questions which I have attached the image......`
one year ago

```							In the given attachment the answer for 1st will be:    sin(a+b)=1So,a+b=90  (in degrees).......(i)    sin(a-b)=1/2So,a-b=60............(ii)By solving the above two equations we geta=60b=30Thus, tan(a+2b)=tan(60+2(30))         =tan(120) =-(3)½Similarly, 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.
```
one year ago
```							Answer for the second question is:           5cosA+3=0       cosA=-3/5    &     sinA=-4/5  (BY PYTHAGORAS THEOREM) tanA=4/3The quadratic euation formed will be:(x-4/5)(x+4/3)=0(5x-4)(3x+4)=015x2 +20x-12x-16=015x2 +8x-16=0 Thus, the answer is (B)15x2+8x-16=0
```
one year 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)1/10xSo, the correct option is (D)
```
one year ago
```							 In the given attachment the answer for 1st will be:    sin(a+b)=1So,a+b=90  (in degrees).......(i)    sin(a-b)=1/2So,a-b=60............(ii)By solving the above two equations we geta=60b=30Thus, tan(a+2b)=tan(60+2(30))         =tan(120) =-(3)½Similarly, 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) tanA=4/3The quadratic euation formed will be:(x-4/5)(x+4/3)=0(5x-4)(3x+4)=015x2 +20x-12x-16=015x2 +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)1/10xSo, the correct option is (D)
```
one year ago
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