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2

Q1. If Prove that

Solution

                                    squaring                 Hence     
Q2. Prove that

Solution

   tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 cos squared space begin display style bevelled x over 2 end style end root plus square root of 2 sin squared space bevelled x over 2 end root over denominator square root of 2 c o s squared space bevelled x over 2 end root minus square root of 2 s i n squared space bevelled x over 2 end root end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator c o s space bevelled x over 2 space plus space sin space bevelled x over 2 over denominator c o s space bevelled x over 2 space minus space s i n space bevelled x over 2 end fraction close parentheses                                                        
Q3. Solve for

Solution

           
Q4. Prove that :sin to the power of negative 1 end exponent 12 over 13 plus cos to the power of negative 1 end exponent 4 over 5 plus tan to the power of negative 1 end exponent 63 over 16 equals straight pi

Solution

We have, sin to the power of negative 1 end exponent 12 over 13 plus cos to the power of negative 1 end exponent 4 over 5 plus tan to the power of negative 1 end exponent 63 over 16 open square brackets because sin to the power of negative 1 end exponent 12 over 13 space equals tan to the power of negative 1 end exponent 12 over 5 space and space cos to the power of negative 1 end exponent 4 over 5 space equals tan to the power of negative 1 end exponent 3 over 4 space close square brackets equals tan to the power of negative 1 end exponent 12 over 5 space plus space tan to the power of negative 1 end exponent 3 over 4 space plus space tan to the power of negative 1 end exponent 63 over 16 equals straight pi plus tan to the power of negative 1 end exponent open curly brackets fraction numerator begin display style 12 over 5 end style plus begin display style 3 over 4 end style over denominator 1 minus begin display style 12 over 5 end style cross times 3 over 4 end fraction close curly brackets plus space tan to the power of negative 1 end exponent 63 over 16 equals straight pi plus tan to the power of negative 1 end exponent fraction numerator negative 63 over denominator 16 end fraction plus tan to the power of negative 1 end exponent 63 over 16 equals straight pi  
Q5. Evaluate

Solution

  Let                                       Applying componend and dividend                         fraction numerator 1 minus tan squared straight x plus 1 plus tan squared straight x over denominator 1 minus tan squared straight x minus 1 minus tan squared straight x end fraction equals fraction numerator 3 plus 5 over denominator 3 minus 5 end fraction minus fraction numerator 1 over denominator tan squared straight x end fraction equals negative 4 tan squared straight x space equals space 1 fourth tanx space equals 1 half                         
Q6. Prove that

Solution

                                                                                
Q7. Find the principal value  of

Solution

We know that sin–1 x denotes an angle in the interval open square brackets negative straight pi over 2 comma straight pi over 2 close square brackets whose sine is x for x element of [–1, 1]. therefore = An angle in  open square brackets negative straight pi over 2 comma straight pi over 2 close square brackets whose sine is   = negative straight pi over 4
Q8. Solve the following equation:  tan–1 2x + tan–1 3x = straight pi over 4

Solution

We have,     tan–1 2x + tan–1 3x = straight pi over 4   = tan–1 1, if  6x2  <  1   = 1, if  6x2  <  1  6x2 + 5x – 1 = 0 and  (6x – 1) (x + 1) = 0 and x =  x =  
Q9. Evaluate 

Solution

 becausedoes not lie between 0 and p. Now,  =                     Þ  = [cos (2p – q) = cos q] Þ  =                                 
Q10. Write

Solution

                       
Q11. Evaluate

Solution

 Let                                
Q12. What is the domain and range of inverse secant function?  Draw its rough sketch.

Solution

    We define inverse secant function as                         as                    domain                          Range =         Graph              



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