1.   Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

2.   Write all the other trigonometric ratios of ∠ A in terms of sec A.

3.   Evaluate :
     (i) ${sin^2   63°+ sin^2   27° }/{cos^2 17° + cos^2   73°}$
	 (ii)   $sin 25° cos 65° + cos 25° sin 65°$
	 
4.   Choose the correct option. Justify your choice. 
	(i)   9 sec2  A – 9 tan2  A =
	(A)  1                        (B)  9                         (C)  8                           (D)  0

	(ii)   (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
	(A)  0                        (B)  1                         (C)  2                           (D)  –1 

	(iii)   (sec A + tan A) (1 – sin A) =

	(A)  sec A                (B)  sin A                  (C)  cosec A                (D)  cos A

	(iv) ${1 + tan^2   A  }/{1 + cot^2 A }$	 

	(A)  sec2  A               (B)  –1           (C)  cot2  A         (D)  tan2  A
	
5.   Prove the following identities, where the angles involved are acute angles for which the expressions are defined. 

	(i) ${(cosec θ – cot θ)^2 ={1 − cos θ}/{1 + cos θ}}$      
	
	(ii) ${cosA}/{(1 + sin A)} + {(1 + sin A) }/{cos A}={2 sec A}$ 
	
	(iii) ${tan θ}/{1− cot θ} + {cot θ}/{1 − tan θ}={1 + sec θcosec θ}$
	
	(iv) ${1 + sec A  }/{sec A }={sin^2A}/{1-cos A}$
	
	(v) ${cos A  – sin  A  + 1  }/{cos A  +  sin  A  – 1}={cosec A  + cot  A}$
	
	(vi) $√{(1 + sin A)/(1 – sin A)} = {secA +tan A}$
	
	(vii) ${   sin θ − 2 sin^3 θ}/{ 2 cos^3 θ – cos θ } =tan θ $
	
	(viii) ${(sin A + cosec A)^2 + (cos A + sec A)^2  = 7 + tan^2 A + cot^2 }$
	
	(ix) $(cosec A  – sin A)(sec A  – cos A)={1}/{tan A  + cot A} $
	
	(x) $({1 +  tan^2  A}/{1 + cot^2A }) $ = $({1 −  tan A }/{1 – cot A })^2$ =$tan^2 A$
	



Solution