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Plzz ans both...................................more than one ootions are correct

Plzz ans both...................................more than one ootions are correct 

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Grade:11

3 Answers

Arun
25750 Points
4 years ago
secA = x + 1/4x 
As 1 + tan^2A = sec^2A 
tan^2A = sec^2A – 1   
Therfore, tan^2A = (x + 1/4x)^2 – 1                
                            
                            = x^2 + 2*x*1/4x + 1/16x^2 – 1               
                            
                            = x^2 + 1/2 + 1/16x2 – 1                
                            
                            = x^2 + 1/16x^2 – 1/2                 
                           
                            = (x – 1/4x)^2 
Therefore, tan^2A = x – 1/4x or tan^2A = - (x – 1/4x) 
Substitute the value of secA and tanA in the given equation secA + tanA 
LHS = secA + tanA         
        
         = x + 1/4x + x – 1/4x         
         
         = 2x        
         
          = RHS 
Or 
LHS = secA + tanA        
        
         =x + 1/4x -x + 1/4x          
         
         = 2/4x          
         
         = 1/2x         
        
         = RHS 
Hence proved.
 
 
venkatesh allam
21 Points
4 years ago
A.   1+\tan ^{\2\ }\theta =\sec ^{2} \theta ;  1+\cot ^{\2\ }\theta =\csc ^{2} \theta;  \sec \theta =1/\cos \theta ; \csc \theta =1/\sin \theta ;   F(X)=6*\cos X\sqrt{\tan ^{\2\ }X\ +1} + 2*\ \sin X\sqrt{\cot ^{\2\ }X\ +1}
F(X) = 6*\cos X *\sec X + 2*\sin X*\csc X
F(X) = 6+2=8
ANS: Option 3
​B. \tan \theta +\sec \theta =\sqrt{(X+\frac{1}{4X})^{2}-1}+X+\frac{1}{4X}
Given \sec \theta = X+\frac{1}{4X}
Substituting \sec \theta value in above equation we get
 
\tan \theta +\sec \theta =\sqrt{(X+\frac{1}{4X})^{2}-1}+X+\frac{1}{4X}
By slove this equation we get
\tan \theta +\sec \theta =X-\frac{1}{4X}+X+\frac{1}{4X}
\tan \theta +\sec \theta =2X
ANS: Option 1
 
TEJA KRISHNA
42 Points
4 years ago
 
secA = x + 1/4x 
As 1 + tan^2A = sec^2A 
tan^2A = sec^2A – 1   
Therfore, tan^2A = (x + 1/4x)^2 – 1                
                            
                            = x^2 + 2*x*1/4x + 1/16x^2 – 1               
                            
                            = x^2 + 1/2 + 1/16x2 – 1                
                            
                            = x^2 + 1/16x^2 – 1/2                 
                           
                            = (x – 1/4x)^2 
Therefore, tan^2A = x – 1/4x or tan^2A = - (x – 1/4x) 
Substitute the value of secA and tanA in the given equation secA + tanA 
LHS = secA + tanA         
        
         = x + 1/4x + x – 1/4x         
         
         = 2x        
         
          = RHS 

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