Question -
Answer -
(i) (cosec θ – cotθ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sinA) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cotθ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + secA)/sec A = sin2A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) ( cos A–sinA+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A= 1+cot2A.
(vii) (sin θ – 2sin3θ)/(2cos3θ-cosθ) = tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
Solution
