In a triangle ABC- let AB-23- BC - 3 and CA - 4- Then the value of A - C B is

With standard notations given : c= √(23), a = 3, b = 4 Now, A+ C B = cos Asin A +cos Csin Ccos Bsin B = b^2+c^2 a^22bc·sin A+a^2+b^2 c^22ab·sin Cc^2+a^2 b^22ac· ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

In a triangle...

Question

In a triangle $A B C$ , let $A B = \sqrt{23},$ $B C = 3$ and $C A = 4$ . Then the value of $\frac{cot A + cot C}{cot B}$ is

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Solution

With standard notations given :
$c = \sqrt{23}, a = 3, b = 4$
Now,
$\frac{cot A + cot C}{cot B}$
$= \frac{\frac{cos A}{sin A} + \frac{cos C}{sin C}}{\frac{cos B}{sin B}}$
$= \frac{\frac{b^{2} + c^{2} - a^{2}}{2 b c \cdot sin A} + \frac{a^{2} + b^{2} - c^{2}}{2 a b \cdot sin C}}{\frac{c^{2} + a^{2} - b^{2}}{2 a c \cdot sin B}}$
$= \frac{\frac{b^{2} + c^{2} - a^{2}}{4 Δ} + \frac{a^{2} + b^{2} - c^{2}}{4 Δ}}{\frac{c^{2} + a^{2} - b^{2}}{4 Δ}}$
$= \frac{2 b^{2}}{a^{2} + c^{2} - b^{2}}$
$= \frac{32}{9 + 23 - 16}$
$= 2$

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In a triangle ABC, let AB=√23, BC=3 and CA=4. Then the value of cotA+cotCcotB is

With standard notations given : c=√23,a=3,b=4 Now, cotA+cotCcotB =cosAsinA+cosCsinCcosBsinB =b2+c2−a22bc⋅sinA+a2+b2−c22ab⋅sinCc2+a2−b22ac⋅sinB =b2+c2−a24Δ+a2+b2−c24Δc2+a2−b24Δ =2b2a2+c2−b2 =329+23−16 =2

In a triangle $A B C$ , let $A B = \sqrt{23},$ $B C = 3$ and $C A = 4$ . Then the value of $\frac{cot A + cot C}{cot B}$ is