Question

Prove that (1-sin+cos)²=2(1+cos)(1-sin)

Answer 1

Answer:

Explanation:

the image below gives you the answer

Answer 2

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• We need to prove that (1-sinA+cosA)²=2(1+cos)(1-sin).

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

LHS:

$\sf \to (1-sinA+cosA)^2$

▣ Simplify using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.

$\sf \to cos^2A-2sincos+sin^2+2cos-2sin+1$

▣ Rearrange the terms.

$\sf \to cos^2A+sin^2-2sincos+2cos-2sin+1$

▣ We know that cos²θ+sin²θ=1.

$\sf \to 1-2sincos+2cos-2sin+1$

$\sf \to 2-2sincos+2cos-2sin$

▣ Now, rearrange the terms.

$\sf \to -2sincos+2cos-2sin+2$

▣ Take 2cos common and 2 common.

$\sf \to 2cos(1-sinA)+2(1 - sinA)$

▣ Take (1-sinA) common.

$\sf \to (1-sinA)(2cos+2)$

▣ Take 2 common.

$\to \sf{\red{2(1-sinA)(cos+1)}}$

LHS=RHS.

HENCE PROVED.

FUNDAMENTAL TRIGONOMETRIC IDENTITIES:

$\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}$

T-RATIOS:

[tex]\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array} [/tex]