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Syllabus

in any triangle abc prove

a cosA+b cosB+c cosC=2a sin B sin C

1.if cos(A+B) sin(C+D)=cos(A-B) sin(C-D), show that cotA cotB cotC = cotD

rnShow that the equation cosec x=4ab/(a+ b)

^{2},(ab>0) is possible if a=b49. Determine maximum value of ( 5sinx - 12cosx ) ( 5cosx + 12sinx ).

if cos(a+b) = 0 then find the value of sin(5a+6b)

If 2 tanα = 3 tanβ , prove that tan (α-β) = sin 2β / 5 - cos 2β.

In a triangle ABC , if angle C is obtuse, then the range of tanAtanB is ?

4

^{tan^2x }- 2^{sec^2x}+1 =0 ,x belongs to [0,20]^{2}AsecA + 3^{1/2}tanA =0prove that:

i) tan(A+B)=2tanA

ii) 2sin B= cos A sin(A+B)

iii)sin B= sin A cos(A+B)

iv) [cot A + cot(A+B)] [cot B -3 cot(2A+B)] =6

A, B, C, Dbe positive angles such thatA + B + C + D= 180°, show that sinA. sinB+ sinC. sinD= sin (B + C) . sin (B + D)^{cos3x} . 81^{sin3x }is^{3}A + cos^{3}A +3sinAcosA - 1 =0 , find x^{3}AcosA= 1/4 + cos^{3}AsinA39. In the formula $2\text{cos}\frac{\text{A}}{2}\text{=\xb1}\sqrt{1+\mathrm{sin}A}\text{\xb1}\sqrt{1-\text{sinA}}\text{}$, find within what limits $\frac{\text{A}}{2}$ must lie when

i) The two positive signs are taken

ii) The two negative signs are takes

iii) First sign negative and seconds sign positive.

^{-1}(x)^{ }+sin^{-1}(2x/(1+x^{2})) is independent of x.cot 7 π /16 + 2cot 3 π /8 + cot 15 π /16

Sin A + cos A=??

^{4}π /16 +cos4 3 π /16 + cos4 5π /16 +cos4 7π /16^{2}θ + sin^{2}α)} │ ≤ k, then value of k....................A + B + C= 2π, prove that${\mathrm{cos}}^{2}B+{\mathrm{cos}}^{2}C-{\mathrm{sin}}^{2}A-2\mathrm{cos}A\mathrm{cos}B\mathrm{cos}C=0$

Kis equal to60. The value of tan9$\xb0$ - tan27$\xb0$ - tan63$\xb0$ + tan81$\xb0$ is

(1) 6

(2) 4

(3) 10

(4) 20

(1) 1 (2) 2 (3) $\frac{3}{4}$ (4) $\frac{1}{4}$

ABC, prove that, $\frac{\mathrm{sin}A+\mathrm{sin}B}{2}\le \mathrm{sin}\left(\frac{A+B}{2}\right)$$56.Ifx+y+z=xyzprovethat\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(i\right)\frac{2x}{1-{x}^{2}}+\frac{2y}{1-{y}^{2}}+\frac{2z}{1-{z}^{2}}=\frac{8xyz}{\left(1-{x}^{2}\right)\left(1-{y}^{2}\right)\left(1-{z}^{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(ii\right)\frac{3x-{x}^{3}}{1-3{x}^{2}}+\frac{3y-{y}^{3}}{1-3{y}^{2}}+\frac{3z-{z}^{3}}{1-3{z}^{2}}=\left(\frac{3x-{x}^{3}}{1-3{x}^{2}}\right)\left(\frac{3y-{y}^{3}}{1-3{y}^{2}}\right)\left(\frac{3z-{z}^{3}}{1-3{z}^{2}}\right)$