(A) 1/√3 (B) √3 (C) 1 (D) 0
(C) 1
As indicated by the inquiry,
![\[cos\text{ }9\propto =\text{ }sin\propto and\text{ }9\propto <90{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aa269189f8a96e45274c8cb4c2b746ec_l3.png "Rendered by QuickLaTeX.com")
for example 9α is an intense point
We realize that,
![\[sin\left( 90{}^\circ -\theta \right)\text{ }=\text{ }cos\text{ }\theta \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-43f8ad94e52addfc58ca19a576e5dd93_l3.png "Rendered by QuickLaTeX.com")
In this way,
![\[cos\text{ }9\propto =\text{ }sin\text{ }\left( 90{}^\circ -\propto \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-205555e47ac0c66122499155ce72d4f6_l3.png "Rendered by QuickLaTeX.com")
Since,
![\[cos\text{ }9\propto =\text{ }sin\left( 90{}^\circ -9\propto \right)\text{ }and\text{ }sin\left( 90{}^\circ -\propto \right)\text{ }=\text{ }sin\propto \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1b059cc7bf28b7ff5c3422b4af0d487c_l3.png "Rendered by QuickLaTeX.com")
Subsequently, sin (90°-9∝) = sin∝
![\[90{}^\circ -9\propto =\propto \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d16dd89f9f78ed7fba1533f6540a8df4_l3.png "Rendered by QuickLaTeX.com")
10∝ = 90°
∝ = 9°
Subbing ∝ = 9° in tan 5∝, we get,
![\[tan\text{ }5\propto =\text{ }tan\text{ }\left( 5\times 9 \right)\text{ }=\text{ }tan\text{ }45{}^\circ =\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-01ef0fac16975e347b136ebe31d27493_l3.png "Rendered by QuickLaTeX.com")
∴, tan 5∝ = 1