Question
If secθ + tanθ = 3+√10, then the value
of sinθ+cosθ isSolution
secθ + tanθ = 3+√10 ............... (i) (sec²θ - tan²θ) = 1 (secθ + tanθ) (secθ - tanθ) = 1 (3+√10) (secθ - tanθ) = 1 secθ - tanθ = 1/(3+√10) = 1/(√10+3) secθ - tanθ = √10-3 ............ (ii) From (i) and (ii) we get, 2secθ = 2√10 secθ = √10 cosθ = 1/√10 sin²θ + cos²θ = 1 sin²θ + 1/10 = 1 sin²θ = 1- 1/10 = 9/10 sinθ = 3/√10 sinθ + cosθ = 3/√10 + 1/√10 = 4/√10 = 4/√10 × √10/√10 = (2√10)/5
12, 24, 72, 288, 1440, ?
120     119     ?     352      2464     2455
...12 3 15 ? 153 756
...70    191    47    216    ?    245
105   107   111   114   ?   127
19.11 × 5.98 + 20.03 × 3.12 – 34.95 + 97.9 × 3.02 =?
75    196    52    221    ?    250
5Â Â Â Â Â Â Â Â Â 18Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 44Â Â Â Â Â Â Â Â Â Â Â 83Â Â Â Â Â Â Â Â Â 135Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 200Â Â Â Â ...
3, 4, 10, 33, 136, ?, 4116
If 2.5 3.5 x 30 124 625
Then, x ² - 1 - 2x = ?
...
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