The Total solution for NCERT class 6-12
Evaluate the following:
(i) tan {2 tan-1 (1/5) – π/4}
(ii) Tan {1/2 sin-1 (3/4)}
(iii) Sin {1/2 cos-1 (4/5)}
(iv) Sin (2 tan -1 2/3) + cos (tan-1 √3)
Answer 1 :
(ii) Given tan {1/2sin-1 (3/4)}
(iii) Given sin {1/2cos-1 (4/5)}
(iv) Given Sin (2tan -1 2/3) + cos (tan-1 √3)
Prove the following results:
(i) 2 sin-1 (3/5) = tan-1 (24/7)
(ii) tan-1 ¼ + tan-1 (2/9) = ½ cos-1 (3/5)= ½ sin-1 (4/5)
(iii) tan-1 (2/3) = ½ tan-1 (12/5)
(iv) tan-1 (1/7) + 2 tan-1 (1/3) = π/4
(v) sin-1 (4/5) + 2 tan-1 (1/3) = π/2
(vi) 2 sin-1 (3/5) – tan-1 (17/31) = π/4
(vii) 2 tan-1 (1/5) + tan-1 (1/8) = tan-1 (4/7)
(viii) 2 tan-1 (3/4) – tan-1 (17/31) =π/4
(ix) 2 tan-1 (1/2) + tan-1 (1/7) = tan-1 (31/17)
(x) 4 tan-1(1/5) – tan-1(1/239) = π/4
Answer 2 :
(i) Given 2 sin-1 (3/5)= tan-1 (24/7)
Hence, proved.
(ii) Given tan-1 ¼+ tan-1 (2/9) = ½ cos-1 (3/5) = ½ sin-1 (4/5)
(iii) Given tan-1 (2/3)= ½ tan-1 (12/5)
(iv) Given tan-1 (1/7)+ 2 tan-1 (1/3) = π/4
(v) Given sin-1 (4/5)+ 2 tan-1 (1/3) = π/2
(vi) Given 2 sin-1 (3/5)– tan-1 (17/31) = π/4
(vii) Given 2 tan-1 (1/5)+ tan-1 (1/8) = tan-1 (4/7)
(viii) Given 2 tan-1 (3/4)– tan-1 (17/31) = π/4
(ix) Given 2 tan-1 (1/2)+ tan-1 (1/7) = tan-1 (31/17)
(x) Given 4 tan-1(1/5)– tan-1(1/239) = π/4
If sin-1 (2a/1 + a2) – cos-1(1 – b2/1+ b2) = tan-1(2x/1 – x2), then prove that x =(a – b)/ (1 + a b)
Answer 3 :
Given sin-1 (2a/1+ a2) – cos-1(1 – b2/1 + b2) = tan-1(2x/1– x2)
Prove that:
(i) tan-1{(1 – x2)/ 2x)} + cot-1{(1 – x2)/2x)} = π/2
(ii) sin {tan-1 (1 – x2)/ 2x) + cos-1 (1– x2)/ (1 + x2)} = 1
Answer 4 :
(i) Given tan-1{(1– x2)/ 2x)} + cot-1{(1 – x2)/ 2x)} = π/2
(ii) Given sin {tan-1 (1– x2)/ 2x) + cos-1 (1 – x2)/ (1 + x2)}
Answer 5 :
Given sin-1 (2a/1+ a2) + sin-1 (2b/ 1+ b2) = 2 tan-1 x
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