Prove that cos90 - A-sin 90 - Atan 90 - A - sin2A-

cos(90 A)=sin(A) ; sin(90 A)=cos(A) ; tan(90 A)=(A) (A)=cos(A)sin(A) LHS cos(90 A)sin(90 A)tan(90 A)=sin(A)cos(A)(A) sin(A)cos(A)(A)= ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

Prove that ...

Question

Prove that $\frac{cos (90 - A) \cdot sin (90 - A)}{tan (90 - A)} = {sin}^{2} A$ .

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Solution

$cos (90 - A) = sin (A)$ ;
$sin (90 - A) = cos (A)$ ;
$tan (90 - A) = cot (A)$
$cot (A) = \frac{cos (A)}{sin (A)}$

LHS
$\frac{cos (90 - A) sin (90 - A)}{tan (90 - A)} = \frac{sin (A) cos (A)}{cot (A)}$

$\frac{sin (A) cos (A)}{cot (A)} = \frac{sin (A) cos (A) sin (A)}{cos (A)}$

$\frac{sin (A) cos (A) sin (A)}{cos (A)} = sin (A) sin (A) = {sin}^{2} (A) = L H S$

RHS= ${sin}^{2} (A)$

$L H S = R H S$
Hence Proved

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Q. Prove the following :
(iv)

\frac{cos (90^{\circ} - A) sin (90^{\circ} - A)}{tan (90^{\circ} - A)} = {sin}^{2} A

Prove the following:

(i) $s i n θ s i n (90^{o} - θ) - c o s θ c o s (90^{o} - θ) = 0$
(ii) $\frac{c o s (90^{o} - θ) s e c (90^{o} - θ) t a n θ}{c o s e c (90^{o}) s i n (90^{o} - θ) c o t (90^{o} - θ)} + \frac{t a n (90^{o} - θ)}{c o t θ} = 2$
(iii) $\frac{t a n (90^{o} - A) c o t A}{c o s e c^{2} A} - c o s^{2} A = 0$
(iv) $\frac{c o s (90^{o} - A) s i n (90^{o} - A)}{t a n (90^{o} - A)} = s i n^{2} A$
(v) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$

\frac{cos (90 - A) sin (90 - A)}{tan (90 - A)}

Q. Prove that:

\frac{cos A}{sin (90^{\circ} - A)} + \frac{sin A}{cos (90^{\circ} - A)} = 2

Q. Prove the following :

\frac{cos (90^{\circ} - θ) sec (90^{\circ} - θ) tan θ}{cosec (90^{\circ} - θ) sin (90^{\circ} - θ) cot (90^{\circ} - θ)} + \frac{tan (90^{\circ} - θ)}{cot θ} = 2

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Prove that cos(90−A)⋅sin(90−A)tan(90−A)=sin2A.

cos(90−A)=sin(A);sin(90−A)=cos(A);tan(90−A)=cot(A)cot(A)=cos(A)sin(A)LHScos(90−A)sin(90−A)tan(90−A)=sin(A)cos(A)cot(A)sin(A)cos(A)cot(A)=sin(A)cos(A)sin(A)cos(A)sin(A)cos(A)sin(A)cos(A)=sin(A)sin(A)=sin2(A)=LHSRHS=sin2(A)LHS=RHSHence Proved

Prove that $\frac{cos (90 - A) \cdot sin (90 - A)}{tan (90 - A)} = {sin}^{2} A$ .