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4.     State whether the following are true or false. Justify your answer.

             (i) \sin (A + B) = \sin A + \sin B
             (ii)  The value of \sin \theta  increases as \theta increases.
             (iii) The value of \cos \theta increases as \theta increases.
             (iv)\sin \theta =\cos \theta  for all values of \theta .
             (v) \cot A is not defined for A=0^{o}

(i) False, Let A = B = Then,    (ii) True, Take  whent   = 0 then zero(0),   = 30 then value of  is 1/2 = 0.5  = 45 then value of  is 0.707   (iii) False,   (iv) False, Let  = 0    (v) True,   (not defined)

3.     If  \tan (A+B)=\sqrt{3} and \tan (A-B)= \frac{1}{\sqrt{3}};  0^{o}<A+B\leq 90^{o};A> B, find A \: and \: B.

Given that, So, ..........(i) therefore, .......(ii) By solving the equation (i) and (ii) we get; and  

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A
 

We need to prove, Taking LHS; Taking RHS; LHS = RHS Hence proved.

2. Choose the correct option and justify your choice :

(iv)\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=

  (A)\: \cos 60^{o}                (B)\: \sin 60^{o}              (C)\: \tan 60^{o}            (D)\: \sin 30^{o}  

Put the value of The correct option is (C)

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (ix)\:(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}

                      [Hint : Simplify LHS and RHS separately]

We need to prove- Taking LHS; Taking RHS; LHS = RHS Hence proved.

2. Choose the correct option and justify your choice :

 (iii)\sin \: 2A=2\: \sin A is true when A =

 (A)0^{o}                (B)\: 30^{o}                (C)\: 45^{o}                (D)\: 60^{o}

The correct option is (A) We know that  So, 

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

    (viii)(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A

           

Given equation, ..................(i) Taking LHS; [since ] Hence proved  

2. Choose the correct option and justify your choice :

  (ii)\: \frac{1-\tan^{2}45^{o}}{1+ \tan^{2}45^{o}}=

   (A)\: \tan \: 90^{o}               (B)\: 1            (C) \: \sin 45^{o}             (D) \: 0

The correct option is (D) We know that  So,

2.  Choose the correct option and justify your choice :

      (i)\, \frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=

   (A)\: \sin 60^{o}                (B)\: \cos 60^{o}             (C)\: \tan 60^{o}            (D)\: \sin 30^{o}

Put the value of tan 30 in the given question- The correct option is (A)    

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (vii)\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta

We need to prove - Taking LHS; [we know the identity ] Hence proved.

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A

We need to prove - Taking LHS; By rationalising the denominator, we get; Hence proved.  

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

     (v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}= \csc A+\cot A , using the identity \csc ^{2}A= 1+\cot ^{2}A

            

            

We need to prove - Dividing the numerator and denominator by , we get; Hence Proved.

5.  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta

           [Hint : Write the expression in terms of \sin \theta and \cos\theta]
 

We need to prove- Taking LHS; By using the identity a3 - b3 =(a - b) (a2 + b2+ab) Hence proved.

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A
 

We need to prove- taking LHS;   = RHS Hence proved.

1.        Evaluate the following :

        (v)\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}

.....................(i) We know the values of- By substituting all these values in equation(i), we get;

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (i) (\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }

We need to prove- Now, taking LHS,                                                                        LHS = RHS Hence proved.

1.     Evaluate the following :

      (iv)\: \frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}

..................(i) It is known that the values of the given trigonometric functions, Put all these values in equation (i), we get;

1.     Evaluate the following :

    (iii)\: \frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}

we know the value of   ,  and , After putting these values 

1.     Evaluate the following :

     (ii)\: 2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}

We know the value of   and   According to question,  

4.    Choose the correct option. Justify your choice.

     (iv) \frac{1+\tan ^{2}A}{1+\cot ^{2}A}=

     (A) \sec ^{2}A   (B) -1  (C) \cot ^{2}A    (D) \tan ^{2}A

The correct option is (D) ..........................eq (i) The above equation can be written as; We know that  therefore,
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