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In any triangle ABC, which is not right angled S cos A . cosec B . cosec C is equal to
- a)1
- b)2
- c)3
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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In any triangle ABC, which is not right angled S cos A . cosec B . cos...
Solution:
Given, S cos A . cosec B . cosec C
Let's use the formula of Sine Rule:
a/Sin A = b/Sin B = c/Sin C = 2R, where R is the circumradius of the triangle ABC
So, we can write the given expression as:
S cos A . cosec B . cosec C = S cos A . (2R/b) . (2R/c)
= 4RS cos A / bc
= 2R(Sin 2A) / bc
= R (Sin 2A / Sin A Sin B Sin C) (Using Sin Rule)
= R (2 Sin A Cos A / 2 Sin A Sin B Sin C)
= 1/2 (Cos A / Sin B Sin C)
Now, we know that Cos A = (b^2 + c^2 - a^2) / 2bc (by Cosine Rule)
Putting this in above expression, we get:
S cos A . cosec B . cosec C = 1/2 [(b^2 + c^2 - a^2) / 2bc] . (1/Sin B) . (1/Sin C)
= [(b^2 + c^2 - a^2) / 4bc Sin B Sin C]
= [(2b^2 + 2c^2 - a^2 - b^2 - c^2) / 4bc Sin B Sin C]
= [(b^2 + c^2 - a^2) / 4bc Sin B Sin C]
= [(2bc Cos A) / 4bc Sin B Sin C] (by Cosine Rule)
= (Cos A / 2 Sin B Sin C)
= 1/2 (Cos A / Sin B Sin C)
= 1/2 [(b^2 + c^2 - a^2) / 2bc Sin B Sin C] (by above derivation)
= 1/2 [(b^2 + c^2 - a^2) / 4R S]
= 1/2 [(2b^2 + 2c^2 - a^2 - b^2 - c^2) / 4R S]
= 1/2 [(a^2 - 2b^2 + 2c^2) / 4R S]
= 1/2 [(a^2 + 2bc Cos A) / 4R S] (by Cosine Rule)
= 1/2 (a / 2R)
= Sin A
Therefore, S cos A . cosec B . cosec C = Sin A
Hence, the correct option is (B) 2.
Given, S cos A . cosec B . cosec C
Let's use the formula of Sine Rule:
a/Sin A = b/Sin B = c/Sin C = 2R, where R is the circumradius of the triangle ABC
So, we can write the given expression as:
S cos A . cosec B . cosec C = S cos A . (2R/b) . (2R/c)
= 4RS cos A / bc
= 2R(Sin 2A) / bc
= R (Sin 2A / Sin A Sin B Sin C) (Using Sin Rule)
= R (2 Sin A Cos A / 2 Sin A Sin B Sin C)
= 1/2 (Cos A / Sin B Sin C)
Now, we know that Cos A = (b^2 + c^2 - a^2) / 2bc (by Cosine Rule)
Putting this in above expression, we get:
S cos A . cosec B . cosec C = 1/2 [(b^2 + c^2 - a^2) / 2bc] . (1/Sin B) . (1/Sin C)
= [(b^2 + c^2 - a^2) / 4bc Sin B Sin C]
= [(2b^2 + 2c^2 - a^2 - b^2 - c^2) / 4bc Sin B Sin C]
= [(b^2 + c^2 - a^2) / 4bc Sin B Sin C]
= [(2bc Cos A) / 4bc Sin B Sin C] (by Cosine Rule)
= (Cos A / 2 Sin B Sin C)
= 1/2 (Cos A / Sin B Sin C)
= 1/2 [(b^2 + c^2 - a^2) / 2bc Sin B Sin C] (by above derivation)
= 1/2 [(b^2 + c^2 - a^2) / 4R S]
= 1/2 [(2b^2 + 2c^2 - a^2 - b^2 - c^2) / 4R S]
= 1/2 [(a^2 - 2b^2 + 2c^2) / 4R S]
= 1/2 [(a^2 + 2bc Cos A) / 4R S] (by Cosine Rule)
= 1/2 (a / 2R)
= Sin A
Therefore, S cos A . cosec B . cosec C = Sin A
Hence, the correct option is (B) 2.
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In any triangle ABC, which is not right angled S cos A . cosec B . cosec C is equal toa)1b)2c)3d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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