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Q. No. 15 : Half yearly Exam 2014-15

208 views · Published 9 October 2016 · 3:12 · Indexed 20 September 2026

Channel: Vishwakarma Classes · 2016 · Education

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15.  a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14


Prove that : square of ( cosec θ – cot θ ) =  1 - cosθ  /1 +  cosθ  

(iii)
tan cot
1 sec cosec
1 cot 1 tan θθ + = + θ θ −θ−θ [Hint : Write the expression in terms of sin θ and cos θ]
(iv)
2 1 sec A sin A sec A 1 – cos A + = [Hint : Simplify LHS and RHS separately]
(v)
cos A – sin A + 1
cosec A + cot A,
cos A + sin A – 1
= using the identity cosec2 A = 1 + cot2 A.
(vi)
1 sinA
sec A + tan A
1 – sin A +
= (vii)
3
3 sin 2 sin
tan
2 cos cos θ − θ
=θ
θ − θ (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(ix)
1
(cosecA –sinA)(secA –cosA)
tanA + cot A
=
[Hint : Simplify LHS and RHS separately

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