prove the following: 4. 2 tan inverse 1 / 2 + tan inverse 1 / 7 = tan inverse 31 / 17
prove the following:

    4.
given to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved.
tan−1{2cos(2sin−121​)}=tan−1{2cos(2×6π​)}=tan−1{2cos(3π​)}=tan−1{2×21​}=tan−1{1}=4π​ `sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°` ....(i) by trigonometric ratios we have `sin 30^@ = 1/2` `cos 45^@ = 1/sqrt2` `tan 30^2 = 1/sqrt3` `sin 90^@ = 1 cos 90^@ = 0 cos 0^@ = 1` by substituting above values in (i), we get `[1/2]^2 . [1/sqrt2]^2 + 4[1/sqrt3]^2 + 1/2[1]^2 - 2[0]^2 + 1/24 [1]^2` `1/4.1/2 + 4/ 1/3 + 1/2 - 0 + 1/24` `1/8 + 4/3 + 1/2 + 1/24 = 48/24 = 2` tan a/2 is equal to dear student,
1+ tan2x= 1+sin2x/cos2x
find common denominator which is cos2x
cos2x/cos2x +
sin2x/cos2x
(cos2x + sin2x)/cos2x
1+cot2x = 1+ cos2x/sin2x
common denominator is sin2x
sin2x/sin2x +
cos2x/sin2x
( sin2x + cos2x)/sin2x
1/(cosx2+ sin2x)/cos2x + 1/(
sin2x + cos2x)/sin2x
multiply the numerator and denominator by the reciprocal of the
bottom fractions to get
cos2x/(cosx2+ sin2x )+
sin2x/( cosx2+ sin2x)
since they now have a common denominator you can add the
tips
(cosx2+ sin2x)/(cosx2+
sin2x)
1/1 = 1

regards consider lhs,lhs = secθ (1 – sinθ) (secθ + tanθ)⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ - ) (secθ + tanθ)we know that = tanθ.⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ – tanθ) (secθ + tanθ)we know that (a + b) (a – b) = a2 – b2.⇒ secθ (1 – sinθ) (secθ + tanθ) = sec2θ – tan2θwe know that sec2θ – tan2θ = 1.∴ secθ (1 – sinθ) (secθ + tanθ) = 1 = rhshence proved. let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, solution: `ch\_(4)(g) + 2o\_(2)(g) to co\_(2)(g) + 2h\_(2)o(g) + "heat. (evolved)"` tw - st < sw < st + tw

4 < sw < 34