Question

plz plz solve this question of trigonometry class10th fast with explanation

Answer 1

$\boxed{\mathsf{ Solution : \: }}\\ \\ \mathsf{ Given, \: }\\ \\\mathtt{ \implies \cos \theta \: + \: \sin \theta \: = \: 1 } \\ \\ \mathsf{ Squaring \: \: both \: \: sides , \: }\\ \\\mathtt{ \implies \: ( \: \cos \theta \: + \: \sin \theta \: )^{2} \: = \: 1^{2} } \\ \\ \mathtt{ \implies \: ( \: \cos^{2} \theta \: + \: \sin ^{2} \theta \: + \: 2 \: \sin \theta \: \cos \theta \: ) \: = \: 1 } \\ \\ \mathtt{ \implies \: ( \: 1 \: + \: 2 \: \sin \theta \: \cos \theta \: ) \: = \: \sin^{2} \theta \: + \: \cos ^{2} \theta \quad \quad \boxed{ \implies \: \sin^{2} \theta \: + \: \cos^{2} \theta \: = \: 1 } } \\ \\ \mathtt{ \implies \: 1 \: = \: \sin ^{2} \theta \: + \: \cos^{2} \theta \: - \: 2 \: \sin \theta \: \cos \theta } \\ \\ \mathtt{ \implies \: 1 \: = \: \cos^{2} \theta \: + \: \sin^{2} \theta \: - \: 2 \: \cos \theta \: \sin \theta } \\ \\ \mathtt{ \implies \: 1 \: = \: ( \: \cos \theta \: - \: \sin \theta \: ) ^{2} \quad \quad \boxed{ \implies \: {a}^{2} \: + \: {b}^{2} \: - \: 2 ab \: = \: ( \: a \: - \: b \: )^{2} } } \\ \\ \mathtt{ \implies \: \sqrt{1} \: = \: \cos \theta \: - \: \sin \theta} \\ \\ \mathtt{ \implies \: \pm1 \: = \: \cos \theta \: - \: \sin \theta } \\ \\ \: \boxed{ \mathtt{ \implies \: \cos \theta \: - \: \sin \theta \: = \: \pm1 }}$

$<b><u><large><marquee direction><large<> Proved !!$

Answer 2

byusing aljebric formulla,

(a+b)^2+(a−b)^2 = 2(a^2+b^2)

(cosθ+sinθ)^2+(cosθ−sinθ)^2=2(sin^2θ+cos^2θ)

⇒12+(cosθ−sinθ)^2=2×1

⇒(cosθ−sinθ)^2=2−1=1

⇒cosθ−sinθ=±1